# Find 2^(2^A) % B

Given two integers A and B, the task is to calculate 22A % B.

Examples:

Input: A = 3, B = 5
Output: 1
223 % 5 = 28 % 5 = 256 % 5 = 1.

Input: A = 10, B = 13
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be efficiently solved by breaking it into sub-problems without overflow of integer by using recursion.

Let F(A, B) = 22A % B.
Now, F(A, B) = 22A % B
= 22 * 2A – 1 % B
= (22A – 1 + 2A – 1) % B
= (22A – 1 * 22A – 1) % B
= (F(A – 1, B) * F(A – 1, B)) % B
Therefore, F(A, B) = (F(A – 1, B) * F(A – 1, B)) % B.
The base case is F(1, B) = 221 % B = 4 % B.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long ` ` `  `// Function to return 2^(2^A) % B ` `ll F(ll A, ll B) ` `{ ` ` `  `    ``// Base case, 2^(2^1) % B = 4 % B ` `    ``if` `(A == 1) ` `        ``return` `(4 % B); ` `    ``else` `    ``{ ` `        ``ll temp =  F(A - 1, B); ` `        ``return` `(temp * temp) % B; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``ll A = 25, B = 50; ` ` `  `    ``// Print 2^(2^A) % B ` `    ``cout << F(A, B); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG  ` `{  ` `    ``// Function to return 2^(2^A) % B  ` `    ``static` `long` `F(``long` `A, ``long` `B)  ` `    ``{  ` `     `  `        ``// Base case, 2^(2^1) % B = 4 % B  ` `        ``if` `(A == ``1``)  ` `            ``return` `(``4` `% B);  ` `        ``else` `        ``{  ` `            ``long` `temp = F(A - ``1``, B);  ` `            ``return` `(temp * temp) % B;  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String args[])  ` `    ``{  ` `        ``long` `A = ``25``, B = ``50``;  ` `     `  `        ``// Print 2^(2^A) % B  ` `        ``System.out.println(F(A, B));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga  `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return 2^(2^A) % B ` `def` `F(A, B): ` ` `  `    ``# Base case, 2^(2^1) % B = 4 % B ` `    ``if` `(A ``=``=` `1``): ` `        ``return` `(``4` `%` `B); ` `    ``else``: ` `        ``temp ``=` `F(A ``-` `1``, B); ` `        ``return` `(temp ``*` `temp) ``%` `B; ` ` `  `# Driver code ` `A ``=` `25``; ` `B ``=` `50``; ` ` `  `# Print 2^(2^A) % B ` `print``(F(A, B)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return 2^(2^A) % B ` `static` `long` `F(``long` `A, ``long` `B) ` `{ ` ` `  `    ``// Base case, 2^(2^1) % B = 4 % B ` `    ``if` `(A == 1) ` `        ``return` `(4 % B); ` `    ``else` `    ``{ ` `        ``long` `temp = F(A - 1, B); ` `        ``return` `(temp * temp) % B; ` `    ``} ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``long` `A = 25, B = 50; ` ` `  `    ``// Print 2^(2^A) % B ` `    ``System.Console.WriteLine(F(A, B)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 `

Output:

```46
```

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