#include <bits/stdc++.h>
using namespace std;
long long sum(long long n) {
  long long x = 0;
  while (n != 0) {
    x += n % 10;
    n /= 10;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s, m, y = 0;
    cin >> n >> s;
    m = n;
    long long a = sum(n);
    if (a <= s) {
      cout << "0" << endl;
      continue;
    } else {
      long long i = 0;
      while (a > s) {
        n /= 10;
        i++;
        n++;
        a = sum(n);
      }
      if (n == 0) {
        y = pow(10, i);
      } else {
        y = n * pow(10, i);
      }
    }
    cout << y - m << endl;
  }
  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;
long long sum(long long n) {
  long long x = 0;
  while (n != 0) {
    x += n % 10;
    n /= 10;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s, m, y = 0;
    cin >> n >> s;
    m = n;
    long long a = sum(n);
    if (a <= s) {
      cout << "0" << endl;
      continue;
    } else {
      long long i = 0;
      while (a > s) {
        n /= 10;
        i++;
        n++;
        a = sum(n);
      }
      if (n == 0) {
        y = pow(10, i);
      } else {
        y = n * pow(10, i);
      }
    }
    cout << y - m << endl;
  }
  return 0;
}

```