#include <bits/stdc++.h>
int main() {
  int t, a, b;
  int eq = 0, eq2 = 0, eq3 = 0;
  scanf("%d", &t);
  for (int h = 0; h < t; ++h) {
    scanf("%d %d", &a, &b);
    if (a % b == 0)
      printf("0\n");
    else if (a < b) {
      eq = b - a;
      printf("%d\n", eq);
    } else {
      eq2 = a / b;
      eq3 = ((eq2 + 1) * b) - a;
      printf("%d\n", eq3);
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
int main() {
  int t, a, b;
  int eq = 0, eq2 = 0, eq3 = 0;
  scanf("%d", &t);
  for (int h = 0; h < t; ++h) {
    scanf("%d %d", &a, &b);
    if (a % b == 0)
      printf("0\n");
    else if (a < b) {
      eq = b - a;
      printf("%d\n", eq);
    } else {
      eq2 = a / b;
      eq3 = ((eq2 + 1) * b) - a;
      printf("%d\n", eq3);
    }
  }
  return 0;
}

```