## 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&lt;=T&lt;=30
1&lt;=A&lt;=1000
1&lt;=B&lt;=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
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
```