#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int i, j, k, temp, m, n, count = 0, flag, a, b, res, val;
    long long int s, sum = 0;
    cin >> n >> s;
    temp = n;
    while (temp != 0) {
      sum += temp % 10;
      temp /= 10;
    }
    if (sum <= s) {
      cout << "0";
    } else {
      res = 0;
      temp = 1;
      for (i = 0; i < 18; i++) {
        val = (n / temp) % 10;
        j = temp * ((10 - val) % 10);
        res += j;
        n += j;
        k = n;
        sum = 0;
        while (k != 0) {
          sum += k % 10;
          k /= 10;
        }
        if (sum <= s) {
          break;
        }
        temp *= 10;
      }
      cout << res;
    }
    cout << endl;
  }
  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 int t;
  cin >> t;
  while (t--) {
    long long int i, j, k, temp, m, n, count = 0, flag, a, b, res, val;
    long long int s, sum = 0;
    cin >> n >> s;
    temp = n;
    while (temp != 0) {
      sum += temp % 10;
      temp /= 10;
    }
    if (sum <= s) {
      cout << "0";
    } else {
      res = 0;
      temp = 1;
      for (i = 0; i < 18; i++) {
        val = (n / temp) % 10;
        j = temp * ((10 - val) % 10);
        res += j;
        n += j;
        k = n;
        sum = 0;
        while (k != 0) {
          sum += k % 10;
          k /= 10;
        }
        if (sum <= s) {
          break;
        }
        temp *= 10;
      }
      cout << res;
    }
    cout << endl;
  }
  return 0;
}

```