#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long cnt = 0, x = n;
    while (x) {
      cnt += x % 10;
      x /= 10;
    }
    long long ans = 0, now = 1;
    while (1) {
      if (cnt <= s) {
        cout << ans << endl;
        break;
      }
      long long p = n % 10;
      if (!p) p = 10;
      ans += (10 - p) * now;
      n += 10 - p;
      n /= 10;
      x = n;
      cnt = 0;
      while (x) {
        cnt += x % 10;
        x /= 10;
      }
      now *= 10;
    }
  }
  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 main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long cnt = 0, x = n;
    while (x) {
      cnt += x % 10;
      x /= 10;
    }
    long long ans = 0, now = 1;
    while (1) {
      if (cnt <= s) {
        cout << ans << endl;
        break;
      }
      long long p = n % 10;
      if (!p) p = 10;
      ans += (10 - p) * now;
      n += 10 - p;
      n /= 10;
      x = n;
      cnt = 0;
      while (x) {
        cnt += x % 10;
        x /= 10;
      }
      now *= 10;
    }
  }
  return 0;
}

```