#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, x, sum = 0, ans = 0;
    cin >> n >> x;
    vector<long long int> v;
    long long int nc = n;
    for (long long int i = 0; nc != 0; i++) {
      v.push_back(nc % 10);
      nc = nc / 10;
    }
    for (long long int i = v.size() - 1; i >= 0; i--) {
      sum += v[i];
    }
    long long int j = 0;
    while (sum > x && j < 18) {
      sum -= v[j] - 1;
      if (j < v.size() - 1) v[j + 1] += 1;
      ans += (pow(10, (j + 1))) - pow(10, j) * v[j];
      v[j] = 0;
      j++;
    }
    cout << ans << endl;
  }
  return (0);
}

### 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() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, x, sum = 0, ans = 0;
    cin >> n >> x;
    vector<long long int> v;
    long long int nc = n;
    for (long long int i = 0; nc != 0; i++) {
      v.push_back(nc % 10);
      nc = nc / 10;
    }
    for (long long int i = v.size() - 1; i >= 0; i--) {
      sum += v[i];
    }
    long long int j = 0;
    while (sum > x && j < 18) {
      sum -= v[j] - 1;
      if (j < v.size() - 1) v[j + 1] += 1;
      ans += (pow(10, (j + 1))) - pow(10, j) * v[j];
      v[j] = 0;
      j++;
    }
    cout << ans << endl;
  }
  return (0);
}

```