#include <bits/stdc++.h>
using namespace std;
int main() {
  int64_t a;
  cin >> a;
  while (a--) {
    string b;
    int64_t c, t = 0;
    cin >> b;
    cin >> c;
    int64_t p[b.size()];
    for (int64_t i = 0; i < b.size(); i++) {
      t += (int64_t)b[i] - '0';
      p[i] = t;
    }
    if (p[b.size() - 1] <= c)
      cout << "0\n";
    else {
      int64_t i, temp = (int64_t)b[b.size() - 1] - '0';
      for (i = 1; i < b.size(); i++) {
        if (p[b.size() - i - 1] < c) {
          cout << (int64_t)(pow(10, i) - temp) << "\n";
          ;
          break;
        } else
          temp += ((int64_t)b[b.size() - i - 1] - '0') * (long long)pow(10, i);
      }
      if (i == b.size()) cout << (int64_t)((int64_t)pow(10, i) - temp) << "\n";
    }
  }
}

### Prompt

Create a solution in CPP for 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() {
  int64_t a;
  cin >> a;
  while (a--) {
    string b;
    int64_t c, t = 0;
    cin >> b;
    cin >> c;
    int64_t p[b.size()];
    for (int64_t i = 0; i < b.size(); i++) {
      t += (int64_t)b[i] - '0';
      p[i] = t;
    }
    if (p[b.size() - 1] <= c)
      cout << "0\n";
    else {
      int64_t i, temp = (int64_t)b[b.size() - 1] - '0';
      for (i = 1; i < b.size(); i++) {
        if (p[b.size() - i - 1] < c) {
          cout << (int64_t)(pow(10, i) - temp) << "\n";
          ;
          break;
        } else
          temp += ((int64_t)b[b.size() - i - 1] - '0') * (long long)pow(10, i);
      }
      if (i == b.size()) cout << (int64_t)((int64_t)pow(10, i) - temp) << "\n";
    }
  }
}

```