## Problem: Collection of pens

Everyone have some habits to collect one thing or the other. Harshit also has the craze to collect pens but in 3 piles. In first pile, he collected A pens and in the second pile, he collected B pens but in the third and the last pile , he think of something different. He decided to collect only the minimum number of pens in third pile so that the sum of pens in the three piles will give him a prime number. Note: there should be atleast one pen in the third pile.

Input:

First line contains the test cases,t. Then T test cases follow. Each line of the test case two space separated values A,B ie. number of pens in first and

second pile respectively.

Output:

Print the minimum number of pens that should be there in the third pile so that sum of all three piles produces a prime number.

Constraints:

1<=T<=30

1<=A<=1000

1<=B<=1000

Example:

Input:

2

1 3

4 3

Output

1

4

Explanation:

In first case,Harshit put one pen in first pile and 3 pens in second pile which give 4 as a sum.So if he adds one pen in the third pile, the sum will yield a prime number ie.5

Therefore the answer is 1.

Similarly for other test case

Python 2.7

from math import * def make_primes(): sieve_eratosthenes = [True]*3001 for i in range(2, int(sqrt(len(sieve_eratosthenes)))): if sieve_eratosthenes[i]: temp = i+i for j in range(temp,len(sieve_eratosthenes),i): sieve_eratosthenes[j] = False result = [] for i in range(2, len(sieve_eratosthenes)): if sieve_eratosthenes[i]: result.append(i) return result if __name__ == '__main__': t = input() primes = make_primes() for _ in range(t): a, b = map(int, raw_input().split()) ab = a+b for prime in primes: if prime > ab: print prime-ab break

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