#include <bits/stdc++.h>
using namespace std;
long long int Pow(long long int n) {
  long long int k = 1;
  for (long long int i = 1; i <= n; i++) k = k * 10;
  return k;
}
long long int check(long long int k) {
  long long int sum = 0;
  while (k > 0) {
    sum += (k % 10);
    k /= 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t, n, s, i;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    long long int N = log10(n) + 1;
    long long int sum = 0;
    long long int v = 0;
    if (check(n) > s) {
      for (i = 1; i <= N; i++) {
        sum = (10 - ((n / Pow(i - 1)) % 10) % 10);
        sum = sum * Pow(i - 1);
        n += sum;
        v += sum;
        if (check(n) <= s) {
          cout << v << "\n";
          break;
        }
      }
    } else
      cout << "0\n";
  }
}

### 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;
long long int Pow(long long int n) {
  long long int k = 1;
  for (long long int i = 1; i <= n; i++) k = k * 10;
  return k;
}
long long int check(long long int k) {
  long long int sum = 0;
  while (k > 0) {
    sum += (k % 10);
    k /= 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t, n, s, i;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    long long int N = log10(n) + 1;
    long long int sum = 0;
    long long int v = 0;
    if (check(n) > s) {
      for (i = 1; i <= N; i++) {
        sum = (10 - ((n / Pow(i - 1)) % 10) % 10);
        sum = sum * Pow(i - 1);
        n += sum;
        v += sum;
        if (check(n) <= s) {
          cout << v << "\n";
          break;
        }
      }
    } else
      cout << "0\n";
  }
}

```