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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int n, s;
    cin >> n >> s;
    vector<int> v;
    long long int x = n, sm = 0;
    while (x) {
      v.push_back(x % 10);
      sm += (x % 10);
      x /= 10;
    }
    if (sm <= s)
      cout << 0 << endl;
    else {
      reverse(v.begin(), v.end());
      if (v[0] < s) {
        sm = 0;
        int i = -1;
        while (sm <= s) {
          i++;
          if (i > v.size()) break;
          sm += v[i];
        }
        if (sm > s) {
          sm -= v[i];
          i--;
          while (sm + 1 > s && i >= 0) {
            sm -= v[i];
            i--;
          }
          v[i]++;
          i++;
          for (; i < v.size(); i++) v[i] = 0;
        }
      } else {
        reverse(v.begin(), v.end());
        v.push_back(1);
        reverse(v.begin(), v.end());
        for (int j = 1; j < v.size(); j++) v[j] = 0;
        sm = 0;
        int i = 0;
        while (sm <= s && i < v.size()) {
          sm += v[i];
          i++;
        }
        if (sm > s) {
          v[i - 1]++;
          for (; i < v.size(); i++) v[i] = 0;
        }
      }
      int co = v.size() - 1;
      long long int ans = 0;
      for (int i = 0; i < v.size(); i++, co++) {
        ans = (10 * ans) + v[i];
      }
      cout << ans - n << endl;
    }
  }
  return 0;
}