#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long a = n;
    long long sum = 0;
    long long ans = 0;
    long long d = 0;
    while (a > 0) {
      a = a / 10;
      d++;
    }
    for (long long i = 0; i < d; i++) {
      sum = 0;
      long long aa = n;
      while (aa > 0) {
        sum += aa % 10;
        aa = aa / 10;
      }
      if (sum > s) {
        long long di = n % 10;
        n = n / 10;
        if (di > 0) {
          sum = sum - di;
          n++;
          long long z = 1;
          for (long long j = 0; j < i; j++) {
            z = z * 10;
          }
          ans += z * (10 - di);
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long a = n;
    long long sum = 0;
    long long ans = 0;
    long long d = 0;
    while (a > 0) {
      a = a / 10;
      d++;
    }
    for (long long i = 0; i < d; i++) {
      sum = 0;
      long long aa = n;
      while (aa > 0) {
        sum += aa % 10;
        aa = aa / 10;
      }
      if (sum > s) {
        long long di = n % 10;
        n = n / 10;
        if (di > 0) {
          sum = sum - di;
          n++;
          long long z = 1;
          for (long long j = 0; j < i; j++) {
            z = z * 10;
          }
          ans += z * (10 - di);
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```