#include <bits/stdc++.h>
using namespace std;
long long T, n, s;
signed main() {
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &n, &s);
    long long S = n, sum = 0;
    while (S) {
      sum += S % 10;
      S /= 10;
    }
    long long Ans = 0, cnt = 1;
    while (sum > s) {
      long long c = n % 10;
      Ans += (10 - c) * cnt;
      cnt *= 10;
      n /= 10;
      n++;
      S = n;
      sum = 0;
      while (S) {
        sum += S % 10;
        S /= 10;
      }
    }
    printf("%lld\n", Ans);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long T, n, s;
signed main() {
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &n, &s);
    long long S = n, sum = 0;
    while (S) {
      sum += S % 10;
      S /= 10;
    }
    long long Ans = 0, cnt = 1;
    while (sum > s) {
      long long c = n % 10;
      Ans += (10 - c) * cnt;
      cnt *= 10;
      n /= 10;
      n++;
      S = n;
      sum = 0;
      while (S) {
        sum += S % 10;
        S /= 10;
      }
    }
    printf("%lld\n", Ans);
  }
  return 0;
}

```