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::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(16);
  int t;
  cin >> t;
  while (t--) {
    unsigned long long int n;
    int s;
    cin >> n >> s;
    unsigned long long int tmp = n;
    vector<int> vect;
    while (tmp) {
      vect.push_back(tmp % 10);
      tmp = tmp / 10;
    }
    int sz = vect.size(), sum = 0;
    for (int i = 0; i < sz; i++) sum += vect[i];
    if (sum <= s)
      cout << 0 << endl;
    else {
      reverse(vect.begin(), vect.end());
      int i = 0;
      while (i < sz && s > vect[i]) {
        s = s - vect[i];
        i++;
      }
      if (i == sz)
        cout << 0 << endl;
      else {
        int rem = sz - i;
        unsigned long long int k = 1;
        for (int j = 0; j < rem; j++) k = k * 10;
        unsigned long long int ans = n % k;
        ans = k - ans;
        cout << ans << endl;
      }
    }
  }
  return 0;
}