#include <bits/stdc++.h>
using namespace std;
int sumOfDigits(long long int n) {
  int sum = 0;
  while (n) {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long int n, s, ans;
    cin >> n >> s;
    ans = 0;
    if (sumOfDigits(n) <= s) ans = 0;
    long long int moves = 0;
    long long int temp;
    long long int i = 1;
    while (sumOfDigits(n) > s) {
      temp = (n / i) % 10;
      moves = ((10 - temp) % 10) * i;
      n += moves;
      ans += moves;
      i *= 10;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 sumOfDigits(long long int n) {
  int sum = 0;
  while (n) {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long int n, s, ans;
    cin >> n >> s;
    ans = 0;
    if (sumOfDigits(n) <= s) ans = 0;
    long long int moves = 0;
    long long int temp;
    long long int i = 1;
    while (sumOfDigits(n) > s) {
      temp = (n / i) % 10;
      moves = ((10 - temp) % 10) * i;
      n += moves;
      ans += moves;
      i *= 10;
    }
    cout << ans << endl;
  }
  return 0;
}

```