#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s, ans = 0;
    cin >> n >> s;
    long long m = 1;
    long long temp;
    temp = n;
    long long sum = 0;
    while (temp != 0) {
      sum += temp % 10;
      temp /= 10;
      if (sum > s) break;
    }
    if (sum <= s) {
      cout << 0 << "\n";
      continue;
    }
    m = 10;
    while (true) {
      temp = n;
      if (temp % m == 0) goto here;
      temp += m - temp % m;
      sum = 0;
      while (temp != 0) {
        sum += temp % 10;
        temp /= 10;
        if (sum > s) goto here;
      }
      if (sum <= s) {
        cout << m - n % m << "\n";
        break;
      }
    here:;
      m *= 10;
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s, ans = 0;
    cin >> n >> s;
    long long m = 1;
    long long temp;
    temp = n;
    long long sum = 0;
    while (temp != 0) {
      sum += temp % 10;
      temp /= 10;
      if (sum > s) break;
    }
    if (sum <= s) {
      cout << 0 << "\n";
      continue;
    }
    m = 10;
    while (true) {
      temp = n;
      if (temp % m == 0) goto here;
      temp += m - temp % m;
      sum = 0;
      while (temp != 0) {
        sum += temp % 10;
        temp /= 10;
        if (sum > s) goto here;
      }
      if (sum <= s) {
        cout << m - n % m << "\n";
        break;
      }
    here:;
      m *= 10;
    }
  }
  return 0;
}

```