#include <bits/stdc++.h>
using namespace std;
int t;
long long n, s;
int gao(long long num) {
  int res = 0;
  while (num) {
    res += num % 10;
    num /= 10;
  }
  return res;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &n, &s);
    long long ans = 0;
    for (int i = 1; i <= 19; i++) {
      if (gao(n) <= s) {
        break;
      }
      long long tmp = n, bit = 1;
      for (int j = 1; j <= i; j++) bit *= 10;
      n = n / bit * bit + bit;
      ans += n - tmp;
    }
    printf("%lld\n", ans);
  }
  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>
using namespace std;
int t;
long long n, s;
int gao(long long num) {
  int res = 0;
  while (num) {
    res += num % 10;
    num /= 10;
  }
  return res;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &n, &s);
    long long ans = 0;
    for (int i = 1; i <= 19; i++) {
      if (gao(n) <= s) {
        break;
      }
      long long tmp = n, bit = 1;
      for (int j = 1; j <= i; j++) bit *= 10;
      n = n / bit * bit + bit;
      ans += n - tmp;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```