#include <bits/stdc++.h>
using namespace std;
vector<int> fr;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, m;
    cin >> n >> m;
    string s = to_string(n);
    double sum = 0;
    for (int i = 0; i < s.length(); i++) {
      sum += s[i] - '0';
    }
    if (sum <= m) {
      cout << "0\n";
    } else {
      int idx = -1;
      int temp = 0;
      for (int i = 0; i < s.length(); i++) {
        temp += s[i] - '0';
        if (temp < m) {
          idx = i;
        } else
          break;
      }
      unsigned long long int val = 0, val1 = 1;
      for (int i = idx + 1; i < s.length(); i++) {
        val = val * 10 + s[i] - '0';
        val1 = val1 * 10;
      }
      unsigned long long int ans = val1 - val;
      cout << ans << "\n";
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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;
vector<int> fr;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, m;
    cin >> n >> m;
    string s = to_string(n);
    double sum = 0;
    for (int i = 0; i < s.length(); i++) {
      sum += s[i] - '0';
    }
    if (sum <= m) {
      cout << "0\n";
    } else {
      int idx = -1;
      int temp = 0;
      for (int i = 0; i < s.length(); i++) {
        temp += s[i] - '0';
        if (temp < m) {
          idx = i;
        } else
          break;
      }
      unsigned long long int val = 0, val1 = 1;
      for (int i = idx + 1; i < s.length(); i++) {
        val = val * 10 + s[i] - '0';
        val1 = val1 * 10;
      }
      unsigned long long int ans = val1 - val;
      cout << ans << "\n";
    }
  }
  return 0;
}

```