#include <bits/stdc++.h>
using namespace std;
int main() {
  long long tc;
  cin >> tc;
  while (tc--) {
    long long k;
    string s;
    cin >> s >> k;
    long long n = s.size();
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += (s[i] - '0');
    }
    if (sum <= k) {
      cout << 0 << endl;
      continue;
    }
    long long bal = 0;
    long long i;
    for (i = 0; i < n; i++) {
      long long x = s[i] - '0';
      if (bal + x >= k) {
        break;
      } else
        bal += x;
    }
    long long boro = 1, choto = 0;
    for (long long j = i; j < n; j++) {
      boro *= 10;
      long long x = s[j] - '0';
      choto += x;
      choto *= 10;
    }
    choto /= 10;
    cout << boro - choto << endl;
  }
}

### Prompt

Generate 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;
int main() {
  long long tc;
  cin >> tc;
  while (tc--) {
    long long k;
    string s;
    cin >> s >> k;
    long long n = s.size();
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += (s[i] - '0');
    }
    if (sum <= k) {
      cout << 0 << endl;
      continue;
    }
    long long bal = 0;
    long long i;
    for (i = 0; i < n; i++) {
      long long x = s[i] - '0';
      if (bal + x >= k) {
        break;
      } else
        bal += x;
    }
    long long boro = 1, choto = 0;
    for (long long j = i; j < n; j++) {
      boro *= 10;
      long long x = s[j] - '0';
      choto += x;
      choto *= 10;
    }
    choto /= 10;
    cout << boro - choto << endl;
  }
}

```