#include <bits/stdc++.h>
using namespace std;
int t;
int m, n;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &m, &n);
    if (n > m)
      printf("%d\n", n - m);
    else if (m == n) {
      printf("0\n");
    } else {
      if (m % n == 0) {
        printf("0\n");
      } else {
        printf("%d\n", (m / n + 1) * n - m);
      }
    }
  }
}

### Prompt

Generate 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;
int t;
int m, n;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &m, &n);
    if (n > m)
      printf("%d\n", n - m);
    else if (m == n) {
      printf("0\n");
    } else {
      if (m % n == 0) {
        printf("0\n");
      } else {
        printf("%d\n", (m / n + 1) * n - m);
      }
    }
  }
}

```