#include <bits/stdc++.h>
using namespace std;
int sum(long long a) {
  long long res = 0;
  while (a > 0) {
    res += a % 10;
    a /= 10;
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    long long n = 0;
    if (sum(a) <= b) {
      cout << 0 << "\n";
      continue;
    }
    long long h = 1;
    for (int i = 0; i < 18; i++) {
      int x = (a / h) % 10;
      long long add = h * ((10 - x) % 10);
      a += add;
      n += add;
      if (sum(a) <= b) {
        break;
      }
      h *= 10;
    }
    cout << n << endl;
  }
  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(long long a) {
  long long res = 0;
  while (a > 0) {
    res += a % 10;
    a /= 10;
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    long long n = 0;
    if (sum(a) <= b) {
      cout << 0 << "\n";
      continue;
    }
    long long h = 1;
    for (int i = 0; i < 18; i++) {
      int x = (a / h) % 10;
      long long add = h * ((10 - x) % 10);
      a += add;
      n += add;
      if (sum(a) <= b) {
        break;
      }
      h *= 10;
    }
    cout << n << endl;
  }
  return 0;
}

```