#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T, n, m, k, p, q, i, j, s, x, y;
  cin >> T;
  while (T--) {
    cin >> n >> s;
    x = 0;
    vector<long long int> a;
    q = 0;
    p = n;
    while (n > 0) {
      j = n % 10;
      q = q + j;
      a.push_back(j);
      n = n / 10;
    }
    m = a.size();
    if (q <= s) {
      cout << "0" << endl;
      continue;
    }
    for (i = 0; i <= m - 1; i++) {
      if (q >= s) {
        x = 1;
        q = q - a[i];
        a[i] = 0;
      } else {
        break;
      }
    }
    if (i == m) {
      a.push_back(1);
      reverse(a.begin(), a.end());
      y = 0;
      for (i = 0; i <= a.size() - 1; i++) {
        y = (10 * y) + a[i];
      }
      cout << y - p << endl;
      continue;
    }
    if (x == 1) {
      a[i]++;
    }
    reverse(a.begin(), a.end());
    y = 0;
    for (i = 0; i <= a.size() - 1; i++) {
      y = (10 * y) + a[i];
    }
    cout << y - p << endl;
  }
}

### 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T, n, m, k, p, q, i, j, s, x, y;
  cin >> T;
  while (T--) {
    cin >> n >> s;
    x = 0;
    vector<long long int> a;
    q = 0;
    p = n;
    while (n > 0) {
      j = n % 10;
      q = q + j;
      a.push_back(j);
      n = n / 10;
    }
    m = a.size();
    if (q <= s) {
      cout << "0" << endl;
      continue;
    }
    for (i = 0; i <= m - 1; i++) {
      if (q >= s) {
        x = 1;
        q = q - a[i];
        a[i] = 0;
      } else {
        break;
      }
    }
    if (i == m) {
      a.push_back(1);
      reverse(a.begin(), a.end());
      y = 0;
      for (i = 0; i <= a.size() - 1; i++) {
        y = (10 * y) + a[i];
      }
      cout << y - p << endl;
      continue;
    }
    if (x == 1) {
      a[i]++;
    }
    reverse(a.begin(), a.end());
    y = 0;
    for (i = 0; i <= a.size() - 1; i++) {
      y = (10 * y) + a[i];
    }
    cout << y - p << endl;
  }
}

```