A-Divisibility Problem

A-Divisibility Problem

You are given two positive integers a and b. In one move you can increase a by 1 (replace a with a+1). Your task is to find the minimum number of moves you need to do in order to make a divisible by b. It is possible, that you have to make 0 moves, as a is already divisible by b. You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers a and b (1≤a,b≤109).

Output

For each test case print the answer — the minimum number of moves you need to do in order to make a divisible by b.

Sample test(s)

Input
5
10 4
13 9
100 13
123 456
92 46

Output
2
5
4
333
0

题目大意：对第二个数据进行变化，找到能够被第一个数据整除的步数即可。注意若正好整除步数即为0。

解题思路：水题，找到能够被整除的步数即可。

#include
#include 
using namespace std;
int main() 
{
	int t;
    cin>>t;
    while (t--) 
	{
        int a,b;
        cin>>a>>b;
        int x=a%b;
        if(x!=0)
        	x=b-x;
        cout<<x<<endl;
    }
    return 0;
}