#include <bits/stdc++.h>
using namespace std;
int Sumdigits(long long x) {
  int ans = 0;
  while (x > 0) {
    ans += x % 10;
    x /= 10;
  }
  return (ans);
}
int main(void) {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int Q, S;
  long long N;
  cin >> Q;
  while (Q--) {
    cin >> N >> S;
    int curr;
    long long mult = 10;
    long long add = 0;
    long long ans = 0;
    do {
      curr = Sumdigits(N);
      if (curr <= S) {
        break;
      }
      add = mult - (N % mult);
      N += add;
      mult *= 10;
      ans += add;
    } while (1);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int Sumdigits(long long x) {
  int ans = 0;
  while (x > 0) {
    ans += x % 10;
    x /= 10;
  }
  return (ans);
}
int main(void) {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int Q, S;
  long long N;
  cin >> Q;
  while (Q--) {
    cin >> N >> S;
    int curr;
    long long mult = 10;
    long long add = 0;
    long long ans = 0;
    do {
      curr = Sumdigits(N);
      if (curr <= S) {
        break;
      }
      add = mult - (N % mult);
      N += add;
      mult *= 10;
      ans += add;
    } while (1);
    cout << ans << endl;
  }
  return 0;
}

```