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;
int digit_sum(long long n) {
  int ans = 0;
  while (n > 0) {
    ans = ans + (n % 10);
    n = n / 10;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    if (digit_sum(n) <= s) {
      cout << 0 << "\n";
    } else {
      string digits = to_string(n);
      int cnt = 0;
      long long ans = 0;
      for (int i = 0; i < digits.size(); i++) {
        if (s > (int)(digits[i] - '0')) {
          s = s - (int)(digits[i] - '0');
          cnt++;
          ans = ans * 10 + (int)(digits[i] - '0');
        } else {
          break;
        }
      }
      ans++;
      for (int i = 0; i < digits.size() - cnt; i++) {
        ans = ans * 10;
      }
      cout << ans - n << "\n";
    }
  }
}