#include <bits/stdc++.h>
using namespace std;
int sum_of_digits(long long n) {
  int ret = 0;
  while (n > 0) {
    ret += n % 10;
    n /= 10;
  }
  return ret;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s, ans = 0;
    cin >> n >> s;
    if (sum_of_digits(n) <= s) {
      cout << ans << endl;
      continue;
    }
    long long digit, temp, p = 1;
    for (int i = 1; i <= 18; i++) {
      digit = (n / p) % 10;
      temp = (10 - digit) * p;
      ans += temp;
      n += temp;
      p *= 10;
      if (sum_of_digits(n) <= s) {
        cout << ans << endl;
        break;
      }
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 sum_of_digits(long long n) {
  int ret = 0;
  while (n > 0) {
    ret += n % 10;
    n /= 10;
  }
  return ret;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s, ans = 0;
    cin >> n >> s;
    if (sum_of_digits(n) <= s) {
      cout << ans << endl;
      continue;
    }
    long long digit, temp, p = 1;
    for (int i = 1; i <= 18; i++) {
      digit = (n / p) % 10;
      temp = (10 - digit) * p;
      ans += temp;
      n += temp;
      p *= 10;
      if (sum_of_digits(n) <= s) {
        cout << ans << endl;
        break;
      }
    }
  }
  return 0;
}

```