#include <bits/stdc++.h>
using namespace std;
const long long int P = 1e9 + 7;
const long long int N = 1e6 + 7;
long long int sum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n, s;
    cin >> n >> s;
    if (sum(n) <= s)
      cout << 0 << "\n";
    else {
      long long int ans = 0, m = 1;
      while (sum(n) > s) {
        long long int d = (n / m) % 10;
        long long int a = ((10 - d) % 10) * m;
        n += a;
        ans += a;
        m *= 10;
      }
      cout << ans << '\n';
    }
  }
  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;
const long long int P = 1e9 + 7;
const long long int N = 1e6 + 7;
long long int sum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n, s;
    cin >> n >> s;
    if (sum(n) <= s)
      cout << 0 << "\n";
    else {
      long long int ans = 0, m = 1;
      while (sum(n) > s) {
        long long int d = (n / m) % 10;
        long long int a = ((10 - d) % 10) * m;
        n += a;
        ans += a;
        m *= 10;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

```