#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 30;
int tc;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    string n;
    int s;
    cin >> n >> s;
    int sum = 0;
    for (char c : n) sum += (c - '0');
    long long ans = 0;
    long long pow = 1;
    for (int i = n.length() - 1; i >= 0 && sum > s; i--) {
      int d = n[i] - '0';
      sum -= d;
      if (n[i] != '0') {
        ans += (10 - d) * pow;
        for (int j = i - 1; j >= 0; j--) {
          if (n[j] == '9') {
            n[j] = '0';
            sum -= 9;
          } else {
            n[j]++;
            sum++;
            break;
          }
        }
      }
      pow *= 10;
    }
    cout << ans << "\n";
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int N = 1e6 + 30;
int tc;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    string n;
    int s;
    cin >> n >> s;
    int sum = 0;
    for (char c : n) sum += (c - '0');
    long long ans = 0;
    long long pow = 1;
    for (int i = n.length() - 1; i >= 0 && sum > s; i--) {
      int d = n[i] - '0';
      sum -= d;
      if (n[i] != '0') {
        ans += (10 - d) * pow;
        for (int j = i - 1; j >= 0; j--) {
          if (n[j] == '9') {
            n[j] = '0';
            sum -= 9;
          } else {
            n[j]++;
            sum++;
            break;
          }
        }
      }
      pow *= 10;
    }
    cout << ans << "\n";
  }
}

```