#include <bits/stdc++.h>
using namespace std;
long long sum(string a) {
  long long z = 0;
  for (int i = 0; i < a.size(); i++) {
    z += a[i] - '0';
  }
  return z;
}
int main() {
  long long T;
  cin >> T;
  while (T--) {
    string a;
    cin >> a;
    long long n;
    cin >> n;
    long long i = 0;
    long long p = 1;
    reverse(a.begin(), a.end());
    a += '0';
    long long ans = 0;
    while (sum(a) > n) {
      if (a[i] == '0') {
        i++;
        p *= 10;
        continue;
      }
      long long z = '9' - a[i];
      z++;
      a[i] = '0';
      a[i + 1]++;
      long long j = i;
      while (a[j] >= '9') {
        a[j] = '0';
        a[j + 1]++;
        j++;
      }
      ans += p * z;
      p *= 10;
      i++;
    }
    cout << ans << endl;
  }
}

### 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;
long long sum(string a) {
  long long z = 0;
  for (int i = 0; i < a.size(); i++) {
    z += a[i] - '0';
  }
  return z;
}
int main() {
  long long T;
  cin >> T;
  while (T--) {
    string a;
    cin >> a;
    long long n;
    cin >> n;
    long long i = 0;
    long long p = 1;
    reverse(a.begin(), a.end());
    a += '0';
    long long ans = 0;
    while (sum(a) > n) {
      if (a[i] == '0') {
        i++;
        p *= 10;
        continue;
      }
      long long z = '9' - a[i];
      z++;
      a[i] = '0';
      a[i + 1]++;
      long long j = i;
      while (a[j] >= '9') {
        a[j] = '0';
        a[j + 1]++;
        j++;
      }
      ans += p * z;
      p *= 10;
      i++;
    }
    cout << ans << endl;
  }
}

```