#include <bits/stdc++.h>
using namespace std;
long long sum(long long n) {
  long long ans = 0;
  while (n) ans += n % 10, n /= 10;
  return ans;
}
long long solve(long long n, long long s) {
  if (sum(n) <= s) return 0;
  int add = 10 - (n % 10);
  return (n % 10 ? (solve(n + add, s) + add) : (solve(n / 10, s) * 10));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    cout << solve(n, s) << '\n';
  }
}

### Prompt

Your task is to create 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;
long long sum(long long n) {
  long long ans = 0;
  while (n) ans += n % 10, n /= 10;
  return ans;
}
long long solve(long long n, long long s) {
  if (sum(n) <= s) return 0;
  int add = 10 - (n % 10);
  return (n % 10 ? (solve(n + add, s) + add) : (solve(n / 10, s) * 10));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    cout << solve(n, s) << '\n';
  }
}

```