#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s, ANS;
    cin >> n >> s;
    long long int i, count = 0, sum = 0, last = 0, j, d = 0, p = 0, re = 0,
                     x = 0, m = 0, c2 = 0;
    for (i = n; i > 0; i = i / 10) {
      re = re * 10 + i % 10;
      sum = sum + i % 10;
      count++;
    }
    if (sum <= s)
      cout << "0" << endl;
    else {
      for (i = re; i > 0; i = i / 10) {
        m = m + i % 10;
        if (m >= s) {
          m = m - i % 10;
          x = x + 1;
          break;
        }
        x = x * 10 + i % 10;
        c2 = c2 + 1;
      }
      ANS = x * pow(10, count - c2) - n;
      cout << ANS << endl;
    }
  }
}

### 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() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s, ANS;
    cin >> n >> s;
    long long int i, count = 0, sum = 0, last = 0, j, d = 0, p = 0, re = 0,
                     x = 0, m = 0, c2 = 0;
    for (i = n; i > 0; i = i / 10) {
      re = re * 10 + i % 10;
      sum = sum + i % 10;
      count++;
    }
    if (sum <= s)
      cout << "0" << endl;
    else {
      for (i = re; i > 0; i = i / 10) {
        m = m + i % 10;
        if (m >= s) {
          m = m - i % 10;
          x = x + 1;
          break;
        }
        x = x * 10 + i % 10;
        c2 = c2 + 1;
      }
      ANS = x * pow(10, count - c2) - n;
      cout << ANS << endl;
    }
  }
}

```