// C++ program to find (a^b) mod m for a large 'a' #include   using namespace std; // utility function to calculate a%m unsigned int aModM(string s unsigned int mod) {  unsigned int number = 0;  for (unsigned int i = 0; i < s.length(); i++)  {  // (s[i]-'0') gives the digit value and form  // the number  number = (number*10 + (s[i] - '0'));  number %= mod;  }  return number; } // Returns find (a^b) % m unsigned int ApowBmodM(string &a unsigned int b  unsigned int m) {  // Find a%m  unsigned int ans = aModM(a m);  unsigned int mul = ans;  // now multiply ans by b-1 times and take  // mod with m  for (unsigned int i=1; i<b; i++)  ans = (ans*mul) % m;  return ans; } // Driver program to run the case int main() {  string a = '987584345091051645734583954832576';  unsigned int b=3 m=11;  cout << ApowBmodM(a b m);  return 0; } 

// Java program to find (a^b) mod m for a large 'a' public class GFG {    // utility function to calculate a%m  static int aModM(String s int mod)  {  int number = 0;  for (int i = 0; i < s.length(); i++)  {    // (s[i]-'0') gives the digit  // value and form the number  number = (number * 10 );  int x = Character.getNumericValue(s.charAt(i));  number = number + x;  number %= mod;  }    return number;  }  // Returns find (a^b) % m  static int ApowBmodM(String a int b int m)  {    // Find a%m  int ans = aModM(a m);  int mul = ans;    // now multiply ans by b-1 times   // and take mod with m  for (int i = 1; i < b; i++)  ans = (ans * mul) % m;    return ans;  }  // Driver code  public static void main(String args[])  {  String a = '987584345091051645734583954832576';  int b = 3 m = 11;  System.out.println(ApowBmodM(a b m));  } } // This code is contributed by Sam007 

# Python program to find (a^b) mod m for a large 'a' def aModM(s mod): number = 0 # convert string s[i] to integer which gives # the digit value and form the number for i in range(len(s)): number = (number*10 + int(s[i])) number = number % m return number # Returns find (a^b) % m def ApowBmodM(a b m): # Find a%m  ans = aModM(a m) mul = ans # now multiply ans by b-1 times and take # mod with m for i in range(1b): ans = (ans*mul) % m return ans # Driver program to run the case a = '987584345091051645734583954832576' b m = 3 11 print (ApowBmodM(a b m)) 

 // C# program to find (a^b) mod m // for a large 'a' using System; class GFG {   // utility function to calculate a%m static int aModM(string s int mod) {  int number = 0;  for (int i = 0; i < s.Length; i++)  {    // (s[i]-'0') gives the digit  // value and form the number  number = (number * 10 );  int x = (int)(s[i] - '0');  number = number + x;  number %= mod;  }  return number; } // Returns find (a^b) % m static int ApowBmodM(string a int b   int m) {    // Find a%m  int ans = aModM(a m);  int mul = ans;  // now multiply ans by b-1 times   // and take mod with m  for (int i = 1; i < b; i++)  ans = (ans * mul) % m;  return ans; } // Driver Code public static void Main() {  string a = '987584345091051645734583954832576';  int b=3 m=11;  Console.Write(ApowBmodM(a b m));   } } // This code is contributed by Sam007   

 // PHP program to find (a^b) // mod m for a large 'a' // utility function to  // calculate a%m function aModM($s $mod) { $number = 0; for ($i = 0; $i < strlen($s); $i++) { // (s[i]-'0') gives the digit // value and form the number $number = ($number * 10 + ($s[$i] - '0')); $number %= $mod; } return $number; } // Returns find (a^b) % m function ApowBmodM($a $b$m) { // Find a%m $ans = aModM($a $m); $mul = $ans; // now multiply ans by  // b-1 times and take // mod with m for ($i = 1; $i < $b; $i++) $ans = ($ans * $mul) % $m; return $ans; } // Driver code $a = '987584345091051645734583954832576'; $b = 3; $m = 11; echo ApowBmodM($a $b $m); return 0; // This code is contributed by nitin mittal. ?> 

<script> // JavaScript program to find (a^b) mod m // for a large 'a' // Utility function to calculate a%m function aModM(s mod) {  let number = 0;  for(let i = 0; i < s.length; i++)  {    // (s[i]-'0') gives the digit  // value and form the number  number = (number * 10 );  let x = (s[i] - '0');  number = number + x;  number %= mod;  }  return number; }   // Returns find (a^b) % m function ApowBmodM(a b m) {    // Find a%m  let ans = aModM(a m);  let mul = ans;    // Now multiply ans by b-1 times   // and take mod with m  for(let i = 1; i < b; i++)  ans = (ans * mul) % m;    return ans; }   // Driver Code let a = '987584345091051645734583954832576'; let b = 3 m = 11; document.write(ApowBmodM(a b m)); // This code is contributed by souravghosh0416 </script> 

// C++ program to find (a^b) mod m for a large 'a' with an // efficient approach. #include    using namespace std; typedef long long ll; // Reduce the number B to a small number // using Fermat Little ll MOD(string num int mod) {  ll res = 0;  for (int i = 0; i < num.length(); i++)  res = (res * 10 + num[i] - '0') % mod;  return res; } ll ModExponent(ll a ll b ll m) {  ll result;  if (a == 0)  return 0;  else if (b == 0)  return 1;  else if (b & 1) {  result = a % m;  result = result * ModExponent(a b - 1 m);  }  else {  result = ModExponent(a b / 2 m);  result = ((result * result) % m + m) % m;  }  return (result % m + m) % m; } int main() {  // String input as b is very large  string a = '987584345091051645734583954832576';  // String input as b is very large  ll b = 3;  ll m = 11;  ll remainderA = MOD(a m);  cout << ModExponent(remainderA b m);  return 0; } 

// Java program to find (a^b) mod m for a large 'a' with an // efficient approach. public class GFG  {    // Reduce the number B to a small number  // using Fermat Little  static long MOD(String num long mod)  {  long res = 0;  for (int i = 0; i < num.length(); i++) {  res = (res * 10 + num.charAt(i) - '0') % mod;  }  return res;  }  // Calculate the ModExponent of the given large number  // 'a'  static long ModExponent(long a long b long m)  {  long result;  if (a == 0) {  return 0;  }  else if (b == 0) {  return 1;  }  else if (b % 2 != 0) {  result = a % m;  result = result * ModExponent(a b - 1 m);  }  else {  result = ModExponent(a b / 2 m);  result = ((result * result) % m + m) % m;  }  return (result % m + m) % m;  }  public static void main(String[] args)  {  // String input as a is very large  String a = '987584345091051645734583954832576';  long b = 3;  long m = 11;  long remainderA = MOD(a m);  System.out.println(ModExponent(remainderA b m));  } } // The code is contributed by Gautam goel (gautamgoel962) 

# Python3 program to find (a^b) mod m # for a large 'a' # Utility function to calculate a%m def MOD(s mod): res = 0 for i in range(len(s)): res = (res * 10 + int(s[i])) % mod return res # Returns find (a^b) % m def ModExponent(a b m): if (a == 0): return 0 elif (b == 0): return 1 elif (b % 2 != 0): result = a % m result = result * ModExponent(a b - 1 m) else: result = ModExponent(a b / 2 m) result = ((result * result) % m + m) % m return (result % m + m) % m # Driver Code a = '987584345091051645734583954832576' b = 3 m = 11 remainderA = MOD(a m) print(ModExponent(remainderA b m)) # This code is contributed by phasing17 

// C# program to find (a^b) mod m for a large 'a' with an // efficient approach. using System; using System.Collections.Generic; public class GFG {  // Reduce the number B to a small number  // using Fermat Little  static long MOD(string num long mod)  {  long res = 0;  for (int i = 0; i < num.Length; i++) {  res = (res * 10 + num[i] - '0') % mod;  }  return res;  }  // Calculate the ModExponent of the given large number  // 'a'  static long ModExponent(long a long b long m)  {  long result;  if (a == 0) {  return 0;  }  else if (b == 0) {  return 1;  }  else if (b % 2 != 0) {  result = a % m;  result = result * ModExponent(a b - 1 m);  }  else {  result = ModExponent(a b / 2 m);  result = ((result * result) % m + m) % m;  }  return (result % m + m) % m;  }  // Driver Code  public static void Main(string[] args)  {  // String input as a is very large  string a = '987584345091051645734583954832576';  long b = 3;  long m = 11;  // Function Call  long remainderA = MOD(a m);  Console.WriteLine(ModExponent(remainderA b m));  } } // The code is contributed by phasing17 

<script> // JavaScript program to find (a^b) mod m // for a large 'a' // Utility function to calculate a%m function MOD(s mod) {    var res = 0;  for (var i = 0; i < s.length; i++) {  res = (res * 10 + (s[i] - '0')) % mod;  }  return res; } // Returns find (a^b) % m function ModExponent(a b m) {    var result;  if (a == 0) {  return 0;  }  else if (b == 0) {  return 1;  }  else if (b % 2 != 0) {  result = a % m;  result = result * ModExponent(a b - 1 m);  }  else {  result = ModExponent(a b / 2 m);  result = ((result * result) % m + m) % m;  }  return (result % m + m) % m; }   // Driver Code let a = '987584345091051645734583954832576'; let b = 3 m = 11; var remainderA = MOD(a m); document.write(ModExponent(remainderA b m)); // This code is contributed by shinjanpatra. </script> 

// C++ program to find (a^b) mod m for a large 'a' #include   using namespace std; typedef long long ll; // utility function to calculate a%m and b%m ll aModM(string s ll mod) {  ll number = 0;  for (ll i = 0; i < s.length(); i++)  {  // (s[i]-'0') gives the digit value and form  // the number  number = (number*10 + (s[i] - '0'));  number %= mod;  }  return number; } // Returns find (a^b) % m ll ApowBmodM(ll x ll yll m) {  ll res=1;  while(y)  {  if(y&1)  res=(res*x)%m;  y=y>>1;  x=((x*x)%m+m)%m;  }  return (res%m+m)%m; } // Driver program to run the case int main() {  string a = '987584345091051645734583954832576';  ll b=3;  ll m=11;  // Find a%m  ll x=aModM(am);  cout << ApowBmodM(xbm);  return 0; } 

// Java program to find (a^b) mod m for a large 'a' import java.util.*; class GFG {  // utility function to calculate a%m and b%m  static long aModM(String s long mod)  {  long number = 0;  for (int i = 0; i < s.length(); i++) {  // (s[i]-'0') gives the digit value and form  // the number  number = (number * 10 + (s.charAt(i) - '0'));  number %= mod;  }  return number;  }  // Returns find (a^b) % m  static long ApowBmodM(long x long y long m)  {  long res = 1;  while (y > 0) {  if ((y & 1) != 0)  res = (res * x) % m;  y = y >> 1;  x = ((x * x) % m + m) % m;  }  return (res % m + m) % m;  }  // Driver program to run the case  public static void main(String[] args)  {  String a = '987584345091051645734583954832576';  long b = 3;  long m = 11;  // Find a%m  long x = aModM(a m);  System.out.println(ApowBmodM(x b m));  } } // This code is contributed by phasing17 

# Python3 program to find (a^b) mod m for a large 'a' # utility function to calculate a%m and b%m def aModM(s mod): number = 0; for i in range(len(s)): # int(s[i]) gives the digit value and form # the number number = (number * 10 + int(s[i])); number %= mod; return number; # Returns find (a^b) % m def ApowBmodM(x y m): res = 1; while (y > 0): if (y & 1): res = (res * x) % m; y = y >> 1; x = ((x * x) % m + m) % m; return (res % m + m) % m; # Driver program to run the case a = '987584345091051645734583954832576'; b = 3; m = 11; # Find a%m x = aModM(a m); print(ApowBmodM(x b m)); # This code is contributed by phasing17 

// C# program to find (a^b) mod m for a large 'a' using System; class GFG {  // utility function to calculate a%m and b%m  static long aModM(string s long mod)  {  long number = 0;  for (int i = 0; i < s.Length; i++)  {  // (s[i]-'0') gives the digit value and form  // the number  number = (number * 10 + (s[i] - '0'));  number %= mod;  }  return number;  }  // Returns find (a^b) % m  static long ApowBmodM(long x long y long m)  {  long res = 1;  while (y > 0) {  if ((y & 1) != 0)  res = (res * x) % m;  y = y >> 1;  x = ((x * x) % m + m) % m;  }  return (res % m + m) % m;  }  // Driver program to run the case  public static void Main(string[] args)  {  string a = '987584345091051645734583954832576';  long b = 3;  long m = 11;  // Find a%m  long x = aModM(a m);  Console.WriteLine(ApowBmodM(x b m));  } } // This code is contributed by phasing17 

// JavaScript program to find (a^b) mod m for a large 'a' // utility function to calculate a%m and b%m function aModM(s mod) {  let number = 0;  for (var i = 0; i < s.length; i++) {  // parseInt(s[i]) gives the digit value and form  // the number  number = (number * 10 + parseInt(s[i]));  number %= mod;  }  return number; } // Returns find (a^b) % m function ApowBmodM(x y m) {  let res = 1;  while (y) {  if (y & 1)  res = (res * x) % m;  y = y >> 1;  x = ((x * x) % m + m) % m;  }  return (res % m + m) % m; } // Driver program to run the case let a = '987584345091051645734583954832576'; let b = 3; let m = 11; // Find a%m let x = aModM(a m); console.log(ApowBmodM(x b m)); // This code is contributed by phasing17 

#include    using namespace std; typedef long long ll; // Reduce the number B to a small number  // using Fermat Little ll MOD(string numint mod) {  ll res=0;    for(int i=0;i<num.length();i++)  res=(res*10+num[i]-'0')%mod;    return res; } ll ModExponent(ll all bll m) {  ll result;  if(a==0)  return 0;  else if(b==0)  return 1;  else if(b&1)  {  result=a%m;  result=result*ModExponent(ab-1m);  }  else{  result=ModExponent(ab/2m);  result=((result%m)*(result%m))%m;  }  return (result%m+m)%m; } int main() {  // String input as b is very large  string a = '987584345091051645734583954832576';  // String input as b is very large  string b = '467687655456765756453454365476765';  ll m = 1000000007;  ll remainderA = MOD(am);  ll remainderB = MOD(bm);    cout << ModExponent(remainderA remainderB m);    return 0; } 

/*package whatever //do not write package name here */ import java.io.*; class GFG {  // Reduce the number B to a small number  // using Fermat Little.  static long MOD(String numint mod)  {  long res = 0;  for(int i = 0; i < num.length(); i++)  {  res = (res * 10 + num.charAt(i) - '0') % mod;  }  return res;  }  static long ModExponent(long along blong m){  long result = 0;  if(a == 0)  return 0;  else if(b == 0)  return 1;  else if((b&1) == 1){  result = a % m;  result = result*ModExponent(a b - 1 m);  }  else{  result = ModExponent(a b/2 m);  result = ((result % m)*(result % m)) % m;  }  return (result % m + m) % m;  }  public static void main (String[] args) {  // String input as b is very large  String a = '987584345091051645734583954832576';  // String input as b is very large  String b = '467687655456765756453454365476765';  int m = 1000000007;  long remainderA = MOD(am);  long remainderB = MOD(bm);  System.out.println(ModExponent(remainderA remainderB m));  } } // This code is contributed by aadityapburujwale 

# Python3 program to implement the approach # Reduce the number B to a small number # using Fermat Little def MOD(num mod): res = 0; for i in range(len(num)): res = (res * 10 + int(num[i])) % mod; return res; def ModExponent(a b m): if (a == 0): return 0; elif (b == 0): return 1; elif (b & 1): result = a % m; result = result * ModExponent(a b - 1 m); else: b = b // 2 result = ModExponent(a b m); result = ((result % m) * (result % m)) % m; return (result % m + m) % m; # String input as b is very large a = '987584345091051645734583954832576'; # String input as b is very large b = '467687655456765756453454365476765'; m = 1000000007; remainderA = (MOD(a m)); remainderB = (MOD(b m)); print(ModExponent(remainderA remainderB m)); # This code is contributed by phasing17 

// C# program to implement the approach using System; using System.Collections.Generic; class GFG {  // Reduce the number B to a small number  // using Fermat Little.  static long MOD(string num int mod)  {  long res = 0;  for (int i = 0; i < num.Length; i++) {  res = (res * 10 + num[i] - '0') % mod;  }  return res;  }  static long ModExponent(long a long b long m)  {  long result = 0;  if (a == 0)  return 0;  else if (b == 0)  return 1;  else if ((b & 1) == 1) {  result = a % m;  result = result * ModExponent(a b - 1 m);  }  else {  result = ModExponent(a b / 2 m);  result = ((result % m) * (result % m)) % m;  }  return (result % m + m) % m;  }  public static void Main(string[] args)  {  // String input as b is very large  string a = '987584345091051645734583954832576';  // String input as b is very large  string b = '467687655456765756453454365476765';  int m = 1000000007;  long remainderA = MOD(a m);  long remainderB = MOD(b m);  Console.WriteLine(  ModExponent(remainderA remainderB m));  } } // This code is contributed by phasing17 

// JavaScript program to implement the approach // Reduce the number B to a small number // using Fermat Little function MOD(num mod) {  let res = 0;  for (var i = 0; i < num.length; i++)  res = (res * 10 + parseInt(num[i])) % mod;  return res; } function ModExponent(a b m) {  let result;  if (a == 0n)  return 0n;  else if (b == 0n)  return 1n;  else if (b & 1n) {  result = a % m;  result = result * ModExponent(a b - 1n m);  }  else {  b = b / 2n - (b % 2n);  result = ModExponent(a b m);  result = ((result % m) * (result % m)) % m;  }  return (result % m + m) % m; } // String input as b is very large let a = '987584345091051645734583954832576'; // String input as b is very large let b = '467687655456765756453454365476765'; let m = 1000000007; let remainderA = BigInt(MOD(a m)); let remainderB = BigInt(MOD(b m)); console.log(ModExponent(remainderA remainderB BigInt(m))); // This code is contributed by phasing17 

// C++ program to find (a^b) mod m for a large 'a' #include    using namespace std; typedef long long ll; // utility function to calculate a%m and b%m ll aModM(string s ll mod) {  ll number = 0;  for (ll i = 0; i < s.length(); i++) {  // (s[i]-'0') gives the digit value and form  // the number  number = (number * 10 + (s[i] - '0'));  number %= mod;  }  return number; } // Returns find (a^b) % m ll ApowBmodM(string& a string& b ll m) {  ll res = 1;  // Find a%m  ll x = aModM(a m);  // Find b%m  ll y = aModM(b m);  while (y) {  if (y & 1)  res = (res * x) % m;  y = y >> 1;  x = ((x % m) * (x % m)) % m;  }  return (res % m + m) % m; } // Driver program to run the case int main() {  string a = '987584345091051645734583954832576';  string b = '467687655456765756453454365476765';  ll m = 1000000007;  cout << ApowBmodM(a b m);  return 0; } 

/*package whatever //do not write package name here */ import java.io.*; class GFG {  // utility function to calculate a%m and b%m  static long aModM(String s long mod){  long number = 0;  for (int i = 0; i < s.length(); i++)  {  // (s.charAt(i)-'0') gives the digit value and form  // the number  number = (number * 10 + (s.charAt(i) - '0'));  number %= mod;  }  return number;  }  // Returns find (a^b) % m  static long ApowBmodM(String a String b long m)  {  long res = 1;  // Find a%m  long x = aModM(a m);  // Find b%m  long y = aModM(b m);  while (y>0) {  if ((y & 1)==1)  res = (res * x) % m;  y = y >> 1;  x = ((x % m) * (x % m)) % m;  }  return (res % m + m) % m;  }  public static void main (String[] args) {  String a = '987584345091051645734583954832576';  String b = '467687655456765756453454365476765';  long m = 1000000007;  System.out.println(ApowBmodM(a b m));  } } // This code is contributed by aadityapburujwale 

# Python3 program to find (a^b) mod m for a large 'a' # utility function to calculate a%m and b%m def aModM(s mod): number = 0 for i in range(len(s)): # (s[i]-'0') gives the digit value and form # the number number = (number * 10 + (int(s[i]))) number %= mod return number # Returns find (a^b) % m def ApowBmodM(a b m): res = 1 # Find a%m x = aModM(a m) # Find b%m y = aModM(b m) while (y > 0): if (y & 1): res = (res * x) % m y = y >> 1 x = ((x % m) * (x % m)) % m return (res % m + m) % m # Driver program to run the case a = '987584345091051645734583954832576' b = '467687655456765756453454365476765' m = 1000000007 print(ApowBmodM(a b m)) # This code is contributed by phasing17 

// JavaScript program to find (a^b) mod m for a large 'a' // utility function to calculate a%m and b%m function aModM(s mod) {  let number = 0n;  for (let i = 0; i < s.length; i++) {  // (s[i]-'0') gives the digit value and form  // the number  number = (number * 10n + BigInt(parseInt(s[i])));  number %= mod;  }  return number; } // Returns find (a^b) % m function ApowBmodM(a b m) {  let res = 1n;  // Find a%m  let x = BigInt(aModM(a m));  // Find b%m  let y = BigInt(aModM(b m));  while (y > 0n) {  if (y & 1n)  res = (res * x) % m;  y = y >> 1n;  x = ((x % m) * (x % m)) % m;  }  return (res % m + m) % m; } // Driver program to run the case let a = '987584345091051645734583954832576'; let b = '467687655456765756453454365476765'; let m = 1000000007n; console.log(ApowBmodM(a b m)); // This code is contributed by phasing17 

// C# program to find (a^b) mod m for a large 'a' using System; using System.Collections.Generic; class GFG {  // utility function to calculate a%m and b%m  static long aModM(string s long mod)  {  long number = 0;  for (int i = 0; i < s.Length; i++) {  // (s[i]-'0') gives the digit value and form  // the number  number = (number * 10 + (s[i] - '0'));  number %= mod;  }  return number;  }  // Returns find (a^b) % m  static long ApowBmodM(string a string b long m)  {  long res = 1;  // Find a%m  long x = aModM(a m);  // Find b%m  long y = aModM(b m);  while (y != 0) {  if ((y & 1) != 0)  res = (res * x) % m;  y = y >> 1;  x = ((x % m) * (x % m)) % m;  }  return (res % m + m) % m;  }  // Driver program to run the case  public static void Main(string[] args)  {  string a = '987584345091051645734583954832576';  string b = '467687655456765756453454365476765';  long m = 1000000007;  Console.WriteLine(ApowBmodM(a b m));  } } // This code is contributed by phasing17