#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    vector<int> d;
    long long temp = n;
    int sum = 0;
    while (temp > 0) {
      int x = temp % 10;
      sum += x;
      d.push_back(x);
      temp /= 10;
    }
    if (sum <= s) {
      cout << 0 << '\n';
      continue;
    }
    long long ans = 0;
    int i = 0;
    while (sum > s - 1 && i < d.size()) {
      sum -= d[i];
      i++;
    }
    ans = (long long)pow(10, i);
    long long r = n % ans;
    ans -= r;
    cout << ans << '\n';
  }
}

### Prompt

Please provide a CPP coded solution 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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    vector<int> d;
    long long temp = n;
    int sum = 0;
    while (temp > 0) {
      int x = temp % 10;
      sum += x;
      d.push_back(x);
      temp /= 10;
    }
    if (sum <= s) {
      cout << 0 << '\n';
      continue;
    }
    long long ans = 0;
    int i = 0;
    while (sum > s - 1 && i < d.size()) {
      sum -= d[i];
      i++;
    }
    ans = (long long)pow(10, i);
    long long r = n % ans;
    ans -= r;
    cout << ans << '\n';
  }
}

```