#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t, n, y;
  vector<long long> v;
  string s;
  set<long long> s1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long a, b;
    cin >> a >> b;
    long long k = a / b;
    if (a % b == 0)
      cout << "0";
    else if (a > b) {
      k = k + 1;
      cout << (b * k) - a;
    } else if (a < b) {
      cout << b - a;
    }
    cout << 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() {
  long long t, n, y;
  vector<long long> v;
  string s;
  set<long long> s1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long a, b;
    cin >> a >> b;
    long long k = a / b;
    if (a % b == 0)
      cout << "0";
    else if (a > b) {
      k = k + 1;
      cout << (b * k) - a;
    } else if (a < b) {
      cout << b - a;
    }
    cout << endl;
  }
}

```