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;
const int maxn = 1e2 + 5, INF = 0x3f3f3f3f;
char s[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int sum;
    long long ans = 0, ans1 = 0;
    cin >> s >> sum;
    int n = strlen(s);
    for (int i = 0; i < n; i++) {
      if (s[i] - '0' + ans < sum)
        ans += s[i] - '0';
      else {
        int cur = i, flag = 0;
        if (s[cur] - '0' + ans == sum) {
          cur++;
          while (cur < n) {
            if (s[cur] != '0') flag = 1;
            cur++;
          }
          if (flag == 0) break;
        }
        long long t = 0, t1 = 1;
        for (int j = i; j <= n - 1; j++) {
          t = t * 10 + s[j] - '0';
          t1 *= 10;
        }
        ans1 = t1 - t;
        break;
      }
    }
    cout << ans1 << endl;
  }
  return 0;
}