Given two integers, **A** and **B**, the task is to find whether it is possible to make **A** equal to **B** if you are allowed to subtract a prime number **P** any number of times from **A**.**Examples:**

Input:A = 10, B = 4Output:YESExplanation:

Let P = 2 and after subtracting it

three times from AInput:A = 41, B = 40Output:NO

**Approach:** The key observation in the problem is we have to represent the number A as

, As we know every number is divisible by some prime number except 1. Therefore if we find the difference of the number

and if the difference is greater than 1 then both the number can be made equal by subtracting a prime number X times from A.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find if ` `// it is possible to make a equal to b ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find if ` `// it is possible to make ` `// A equal to B ` `bool` `isPossible(` `int` `A, ` `int` `B) ` `{ ` ` ` `return` `(A - B > 1); ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `A = 10, B = 4; ` ` ` ` ` `// Function Call ` ` ` `if` `(isPossible(A, B)) ` ` ` `cout << ` `"Yes"` `; ` ` ` `else` ` ` `cout << ` `"No"` `; ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation to find if ` `// it is possible to make a equal to b ` `class` `GFG{ ` ` ` `// Function to find if ` `// it is possible to make ` `// A equal to B ` `static` `boolean` `isPossible(` `int` `A, ` `int` `B) ` `{ ` ` ` `return` `(A - B > ` `1` `); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `int` `A = ` `10` `, B = ` `4` `; ` ` ` ` ` `// Function Call ` ` ` `if` `(isPossible(A, B)) ` ` ` `System.out.print(` `"Yes"` `); ` ` ` `else` ` ` `System.out.print(` `"No"` `); ` `} ` `} ` ` ` `// This code is contributed by shivanisinghss2110 ` |

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## Python3

`# Python3 implementation to find if ` `# it is possible to make a equal to b ` ` ` `# Function to find if ` `# it is possible to make ` `# A equal to B ` `def` `isPossible(A, B): ` ` ` ` ` `return` `(A ` `-` `B > ` `1` `); ` ` ` `# Driver Code ` `A ` `=` `10` `; B ` `=` `4` `; ` ` ` `# Function Call ` `if` `(isPossible(A, B)): ` ` ` `print` `(` `"Yes"` `); ` `else` `: ` ` ` `print` `(` `"No"` `); ` ` ` `# This code is contributed by Code_Mech` |

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## C#

`// C# implementation to find if ` `// it is possible to make a equal to b ` `using` `System; ` `class` `GFG{ ` ` ` `// Function to find if ` `// it is possible to make ` `// A equal to B ` `static` `bool` `isPossible(` `int` `A, ` `int` `B) ` `{ ` ` ` `return` `(A - B > 1); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main() ` `{ ` ` ` `int` `A = 10, B = 4; ` ` ` ` ` `// Function Call ` ` ` `if` `(isPossible(A, B)) ` ` ` `Console.Write(` `"Yes"` `); ` ` ` `else` ` ` `Console.Write(` `"No"` `); ` `} ` `} ` ` ` `// This code is contributed by Code_Mech ` |

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**Output:**

Yes

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