#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
void solve() {
  long long int n, i, m = 0, q = 0, mx = 1000000000000000,
                      mn = 1000000000000000, k = 0, j = 0, c = 0, d = 0, t = 0,
                      sm = 0, pdt = 1;
  string str;
  cin >> str;
  cin >> m;
  n = str.length();
  long long int cnt = 0;
  long long int mult = 1;
  for (i = 0; i < n; i++) {
    sm += str[i] - '0';
  }
  if (sm <= m) {
    cout << 0 << endl;
    return;
  }
  i = n - 1;
  while (i >= 0) {
    cnt += (10 - (str[i] - '0')) * mult;
    mult *= 10;
    str[i - 1]++;
    sm++;
    sm -= (str[i] - '0');
    if (sm <= m) {
      break;
    }
    i--;
  }
  cout << cnt << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
using namespace std::chrono;
void solve() {
  long long int n, i, m = 0, q = 0, mx = 1000000000000000,
                      mn = 1000000000000000, k = 0, j = 0, c = 0, d = 0, t = 0,
                      sm = 0, pdt = 1;
  string str;
  cin >> str;
  cin >> m;
  n = str.length();
  long long int cnt = 0;
  long long int mult = 1;
  for (i = 0; i < n; i++) {
    sm += str[i] - '0';
  }
  if (sm <= m) {
    cout << 0 << endl;
    return;
  }
  i = n - 1;
  while (i >= 0) {
    cnt += (10 - (str[i] - '0')) * mult;
    mult *= 10;
    str[i - 1]++;
    sm++;
    sm -= (str[i] - '0');
    if (sm <= m) {
      break;
    }
    i--;
  }
  cout << cnt << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```