#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
const long long INF = 1e18;
using namespace std;
long long sum(long long num) {
  long long ans = 0;
  while (num > 0) {
    ans += (num % 10);
    num /= 10;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, second, ans = 0, cur = 1;
    cin >> n >> second;
    if (sum(n) <= second)
      cout << 0 << endl;
    else {
      while (sum(n) > second) {
        ans += ((10 - n % 10) % 10) * cur;
        cur *= 10;
        n += ((10 - n % 10) % 10);
        n /= 10;
      }
      cout << ans << endl;
    }
  }
}

### 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>
const int MOD = 1e9 + 7;
const long long INF = 1e18;
using namespace std;
long long sum(long long num) {
  long long ans = 0;
  while (num > 0) {
    ans += (num % 10);
    num /= 10;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, second, ans = 0, cur = 1;
    cin >> n >> second;
    if (sum(n) <= second)
      cout << 0 << endl;
    else {
      while (sum(n) > second) {
        ans += ((10 - n % 10) % 10) * cur;
        cur *= 10;
        n += ((10 - n % 10) % 10);
        n /= 10;
      }
      cout << ans << endl;
    }
  }
}

```