#include <bits/stdc++.h>
using namespace std;
long long T, n, s, k;
long long ssum(long long k) {
  long long tot = 0;
  while (k) {
    tot += k % 10;
    k /= 10;
  }
  return tot;
}
int main() {
  cin >> T;
  while (T--) {
    cin >> n >> s;
    k = 0;
    int now = ssum(n);
    if (now <= s) {
      cout << 0 << endl;
      continue;
    } else {
      long long tot = 1;
      while (now > s) {
        tot *= 10;
        if (!(n % tot)) continue;
        k += tot - (n % tot);
        n += tot - (n % tot);
        now = ssum(n);
      }
      cout << k << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long T, n, s, k;
long long ssum(long long k) {
  long long tot = 0;
  while (k) {
    tot += k % 10;
    k /= 10;
  }
  return tot;
}
int main() {
  cin >> T;
  while (T--) {
    cin >> n >> s;
    k = 0;
    int now = ssum(n);
    if (now <= s) {
      cout << 0 << endl;
      continue;
    } else {
      long long tot = 1;
      while (now > s) {
        tot *= 10;
        if (!(n % tot)) continue;
        k += tot - (n % tot);
        n += tot - (n % tot);
        now = ssum(n);
      }
      cout << k << endl;
    }
  }
  return 0;
}

```