#include <bits/stdc++.h>
using namespace std;
long long n, s, sum, p;
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    sum = n;
    p = 0;
    while (sum) {
      p = p + sum % 10;
      sum = sum / 10;
    }
    if (p <= s) {
      cout << 0 << "\n";
      continue;
    }
    long long r = 10;
    sum = n;
    long long x, q;
    long long yt = n;
    while (p > s) {
      x = n % r;
      q = r - x;
      n = n + q;
      p = 0;
      yt = n;
      while (yt) {
        p = p + yt % 10;
        yt = yt / 10;
      }
      r = r * 10;
    }
    cout << n - sum << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 n, s, sum, p;
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    sum = n;
    p = 0;
    while (sum) {
      p = p + sum % 10;
      sum = sum / 10;
    }
    if (p <= s) {
      cout << 0 << "\n";
      continue;
    }
    long long r = 10;
    sum = n;
    long long x, q;
    long long yt = n;
    while (p > s) {
      x = n % r;
      q = r - x;
      n = n + q;
      p = 0;
      yt = n;
      while (yt) {
        p = p + yt % 10;
        yt = yt / 10;
      }
      r = r * 10;
    }
    cout << n - sum << "\n";
  }
  return 0;
}

```