#include <bits/stdc++.h>
using namespace std;
void solve() {
  string n;
  int s;
  cin >> n >> s;
  int sum = 0;
  for (int i = 0; i < n.length(); i++) {
    sum += n[i] - 48;
  }
  if (sum <= s) {
    cout << 0 << endl;
    return;
  }
  int count = 1;
  int pos = 0;
  for (int i = 0; i < n.length(); i++) {
    pos = i;
    if (count + n[i] - 48 > s) {
      break;
    }
    count += n[i] - 48;
  }
  long long int ans = 0;
  for (int i = pos; i < n.length(); i++) {
    ans = ans * 10 + (57 - n[i]);
  }
  ans++;
  cout << ans << endl;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}

### 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>
using namespace std;
void solve() {
  string n;
  int s;
  cin >> n >> s;
  int sum = 0;
  for (int i = 0; i < n.length(); i++) {
    sum += n[i] - 48;
  }
  if (sum <= s) {
    cout << 0 << endl;
    return;
  }
  int count = 1;
  int pos = 0;
  for (int i = 0; i < n.length(); i++) {
    pos = i;
    if (count + n[i] - 48 > s) {
      break;
    }
    count += n[i] - 48;
  }
  long long int ans = 0;
  for (int i = pos; i < n.length(); i++) {
    ans = ans * 10 + (57 - n[i]);
  }
  ans++;
  cout << ans << endl;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}

```