#include <bits/stdc++.h>
using namespace std;
inline long long sum_of_digits(long long n) {
  long long sum = 0;
  while (n > 0) sum += n % 10, n /= 10;
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tests;
  cin >> tests;
  while (tests--) {
    long long N, S;
    cin >> N >> S;
    long long M = N;
    long long denom = 10;
    while (sum_of_digits(M) > S) {
      M = (M / denom + 1);
      if (M % 10 == 0) {
        denom *= 10;
        M /= 10;
      }
      M *= denom;
    }
    cout << M - N << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
inline long long sum_of_digits(long long n) {
  long long sum = 0;
  while (n > 0) sum += n % 10, n /= 10;
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tests;
  cin >> tests;
  while (tests--) {
    long long N, S;
    cin >> N >> S;
    long long M = N;
    long long denom = 10;
    while (sum_of_digits(M) > S) {
      M = (M / denom + 1);
      if (M % 10 == 0) {
        denom *= 10;
        M /= 10;
      }
      M *= denom;
    }
    cout << M - N << "\n";
  }
  return 0;
}

```