#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    vector<long long> v;
    long long sum = 0;
    long long cp = a;
    while (cp) {
      sum += cp % 10;
      v.push_back(cp % 10);
      cp /= 10;
    }
    v.push_back(0);
    long long res = 0;
    int i = 0;
    while (sum > b) {
      if (v[i] == 0) {
        i++;
        continue;
      }
      long long x1 = round(pow(10, i));
      res += x1 * (long long)(10 - v[i]);
      sum -= v[i] % 10;
      int j = i + 1;
      while (j < v.size()) {
        sum -= v[j];
        v[j] = (v[j] + 1) % 10;
        sum += v[j];
        if (v[j] != 0) break;
        j++;
      }
      i++;
    }
    cout << res << endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    vector<long long> v;
    long long sum = 0;
    long long cp = a;
    while (cp) {
      sum += cp % 10;
      v.push_back(cp % 10);
      cp /= 10;
    }
    v.push_back(0);
    long long res = 0;
    int i = 0;
    while (sum > b) {
      if (v[i] == 0) {
        i++;
        continue;
      }
      long long x1 = round(pow(10, i));
      res += x1 * (long long)(10 - v[i]);
      sum -= v[i] % 10;
      int j = i + 1;
      while (j < v.size()) {
        sum -= v[j];
        v[j] = (v[j] + 1) % 10;
        sum += v[j];
        if (v[j] != 0) break;
        j++;
      }
      i++;
    }
    cout << res << endl;
  }
}

```