#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s, y = 0, i, sum = 0;
    cin >> n >> s;
    string x;
    x = to_string(n);
    for (i = 0; i < x.length(); i++) {
      sum += (x[i] - '0');
    }
    if (sum <= s) {
      cout << "0\n";
      continue;
    } else {
      for (i = x.length() - 1; i >= 0; i--) {
        if (sum <= s) {
          break;
        }
        if (x[i] != '0') {
          y += (10 - (x[i] - '0')) * pow(10, x.length() - i - 1);
          sum -= (x[i] - '0');
          x[i] = '0';
          while (i > 0 && x[i - 1] == '9') {
            x[i - 1] = '0';
            sum -= 9;
            i--;
          }
          if (i > 0)
            x[i - 1]++;
          else
            x.insert(x.begin(), '1');
          sum++;
        }
      }
      long long int a, b;
      a = stoll(x);
      y = a - n;
      cout << y << "\n";
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s, y = 0, i, sum = 0;
    cin >> n >> s;
    string x;
    x = to_string(n);
    for (i = 0; i < x.length(); i++) {
      sum += (x[i] - '0');
    }
    if (sum <= s) {
      cout << "0\n";
      continue;
    } else {
      for (i = x.length() - 1; i >= 0; i--) {
        if (sum <= s) {
          break;
        }
        if (x[i] != '0') {
          y += (10 - (x[i] - '0')) * pow(10, x.length() - i - 1);
          sum -= (x[i] - '0');
          x[i] = '0';
          while (i > 0 && x[i - 1] == '9') {
            x[i - 1] = '0';
            sum -= 9;
            i--;
          }
          if (i > 0)
            x[i - 1]++;
          else
            x.insert(x.begin(), '1');
          sum++;
        }
      }
      long long int a, b;
      a = stoll(x);
      y = a - n;
      cout << y << "\n";
    }
  }
  return 0;
}

```