#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  unsigned long long t;
  cin >> t;
  while (t--) {
    unsigned long long n, s;
    cin >> n >> s;
    string e = to_string(n);
    for (unsigned long long i = 0; i < e.size(); i++) e[i] -= '0';
    unsigned long long sum = 0, t = 0;
    unsigned long long ans =
        (t + 1) * (unsigned long long)powl(10, e.size()) - n;
    for (unsigned long long i = 0; i < e.size(); i++) {
      t = t * 10 + e[i];
      sum += e[i];
      if (sum < s)
        ans = (t + 1) * (unsigned long long)powl(10, e.size() - 1 - i) - n;
    }
    if (sum <= s) ans = 0;
    cout << ans << '\n';
  }
}

### Prompt

Create a solution in CPP for 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;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  unsigned long long t;
  cin >> t;
  while (t--) {
    unsigned long long n, s;
    cin >> n >> s;
    string e = to_string(n);
    for (unsigned long long i = 0; i < e.size(); i++) e[i] -= '0';
    unsigned long long sum = 0, t = 0;
    unsigned long long ans =
        (t + 1) * (unsigned long long)powl(10, e.size()) - n;
    for (unsigned long long i = 0; i < e.size(); i++) {
      t = t * 10 + e[i];
      sum += e[i];
      if (sum < s)
        ans = (t + 1) * (unsigned long long)powl(10, e.size() - 1 - i) - n;
    }
    if (sum <= s) ans = 0;
    cout << ans << '\n';
  }
}

```