#include <bits/stdc++.h>
using namespace std;
long long sumdig(long long n) {
  long long ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    if (sumdig(n) <= s) {
      cout << "0\n";
      continue;
    }
    long long x = n;
    long long p = 10;
    while (x) {
      long long right = n % p;
      x /= 10;
      long long y = sumdig(x + 1);
      if (y <= s) {
        cout << p - right << '\n';
        break;
      }
      p *= 10LL;
    }
  }
}

### 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;
long long sumdig(long long n) {
  long long ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    if (sumdig(n) <= s) {
      cout << "0\n";
      continue;
    }
    long long x = n;
    long long p = 10;
    while (x) {
      long long right = n % p;
      x /= 10;
      long long y = sumdig(x + 1);
      if (y <= s) {
        cout << p - right << '\n';
        break;
      }
      p *= 10LL;
    }
  }
}

```