#include <bits/stdc++.h>
using namespace std;
unsigned long long sumDigits(unsigned long long n) {
  unsigned long long sum = 0;
  while (n) {
    sum += (n % 10);
    n /= 10;
  }
  return sum;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unsigned long long test_cases;
  cin >> test_cases;
  while (test_cases--) {
    unsigned long long n, s;
    cin >> n >> s;
    if (sumDigits(n) <= s) {
      cout << 0 << endl;
    } else {
      unsigned long long ans = 0, exp = 1;
      while (sumDigits(n) > s) {
        unsigned long long digit = (n / exp) % 10;
        unsigned long long add = exp * ((10 - digit) % 10);
        n += add;
        ans += add;
        exp *= 10;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

### 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;
unsigned long long sumDigits(unsigned long long n) {
  unsigned long long sum = 0;
  while (n) {
    sum += (n % 10);
    n /= 10;
  }
  return sum;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unsigned long long test_cases;
  cin >> test_cases;
  while (test_cases--) {
    unsigned long long n, s;
    cin >> n >> s;
    if (sumDigits(n) <= s) {
      cout << 0 << endl;
    } else {
      unsigned long long ans = 0, exp = 1;
      while (sumDigits(n) > s) {
        unsigned long long digit = (n / exp) % 10;
        unsigned long long add = exp * ((10 - digit) % 10);
        n += add;
        ans += add;
        exp *= 10;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```