Problem: You are given a positive integer n. In one move, you can increase n by one (i.e. make n := n + 1). Your task is to find the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers n and s (1 ≤ n ≤ 10^{18}; 1 ≤ s ≤ 162).

Output

For each test case, print the answer: the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

Example

Input


5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1


Output


8
0
500
2128012501878
899999999999999999

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long dsum(long long x) {
  if (!x) return 0;
  return (x % 10) + dsum(x / 10);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long ans = 0;
    long long mul = 10;
    while (dsum(n) > s) {
      ans += mul - (n % mul);
      n += mul - (n % mul);
      mul *= 10;
    }
    cout << ans << endl;
  }
}