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);
  int t;
  cin >> t;
  while (t--) {
    string n;
    int s;
    cin >> n >> s;
    reverse(n.begin(), n.end());
    char temp[] = "0";
    n.insert(n.size(), temp);
    int sum = 0;
    for (int i = 0; i < n.size(); i++) sum += n[i] - 48;
    char ans[25];
    for (int i = 0; i < 25; i++) ans[i] = 0;
    ans[0] = '0';
    int i = 0;
    while (sum > s) {
      int dig = (n[i] - 48);
      int car = 0;
      sum -= dig;
      n[i] = '0';
      if (dig > 0) {
        ans[i] = 10 - dig + 48;
        car = 1;
      } else
        ans[i] = '0';
      for (int j = i + 1; car; j++) {
        int t = (n[j] - 48 + car);
        if (t == 10)
          sum -= n[j] - 48;
        else
          sum++;
        n[j] = (t % 10) + 48;
        car = t / 10;
      }
      i++;
    }
    int size = 0;
    while (ans[++size])
      ;
    reverse(ans, ans + size);
    cout << ans << "\n";
  }
}