#include <bits/stdc++.h>
int main() {
  std::ios_base::sync_with_stdio(false);
  long t;
  std::cin >> t;
  while (t--) {
    std::string num;
    std::cin >> num;
    long s;
    std::cin >> s;
    long cs(0);
    for (long p = 0; p < num.size(); p++) {
      cs += (num[p] - '0');
    }
    if (cs <= s) {
      puts("0");
      continue;
    }
    long long mult(1), cnt(0), carry(0);
    for (long p = num.size() - 1; p >= 0 && cs >= s; p--) {
      cs -= (num[p] - '0');
      long long diff = (10 - (num[p] - '0') - carry) % 10;
      cnt += mult * diff;
      if (num[p] > '0') {
        carry = 1;
      }
      mult *= 10;
    }
    printf("%lld\n", cnt);
  }
  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>
int main() {
  std::ios_base::sync_with_stdio(false);
  long t;
  std::cin >> t;
  while (t--) {
    std::string num;
    std::cin >> num;
    long s;
    std::cin >> s;
    long cs(0);
    for (long p = 0; p < num.size(); p++) {
      cs += (num[p] - '0');
    }
    if (cs <= s) {
      puts("0");
      continue;
    }
    long long mult(1), cnt(0), carry(0);
    for (long p = num.size() - 1; p >= 0 && cs >= s; p--) {
      cs -= (num[p] - '0');
      long long diff = (10 - (num[p] - '0') - carry) % 10;
      cnt += mult * diff;
      if (num[p] > '0') {
        carry = 1;
      }
      mult *= 10;
    }
    printf("%lld\n", cnt);
  }
  return 0;
}

```