#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long int digit_sum(long long int n) {
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int s;
    long long int n;
    cin >> n >> s;
    long long int sum = digit_sum(n);
    long long int divider = 10;
    long long int cost = 0;
    while (sum > s) {
      long long int value = n % divider;
      if (value == 0) {
        divider *= 10;
        continue;
      }
      cost += divider - value;
      n += divider - value;
      sum = digit_sum(n);
      if (divider != 1000000000000000000) {
        divider *= 10;
      }
    }
    cout << cost << endl;
  }
}

### Prompt

Please formulate 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long int digit_sum(long long int n) {
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int s;
    long long int n;
    cin >> n >> s;
    long long int sum = digit_sum(n);
    long long int divider = 10;
    long long int cost = 0;
    while (sum > s) {
      long long int value = n % divider;
      if (value == 0) {
        divider *= 10;
        continue;
      }
      cost += divider - value;
      n += divider - value;
      sum = digit_sum(n);
      if (divider != 1000000000000000000) {
        divider *= 10;
      }
    }
    cout << cost << endl;
  }
}

```