#include <bits/stdc++.h>
using namespace std;
long long int pp(int x, int y) {
  long long int ans = 1;
  for (int i = 1; i <= y; i++) {
    ans = ans * x;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    int s;
    cin >> n >> s;
    vector<int> v;
    int sum = 0;
    long long int h = n;
    while (h / 10 != 0) {
      int x = h % 10;
      v.push_back(x);
      sum += x;
      h = h / 10;
    }
    int x = h % 10;
    v.push_back(x);
    sum += x;
    if (sum <= s) {
      cout << "0" << endl;
    } else {
      reverse(v.begin(), v.end());
      int i = 0;
      int ss = 0;
      int l = v.size();
      while (i < l) {
        ss += v[i];
        if (ss < s)
          i++;
        else
          break;
      }
      if (i == 0) {
        cout << pp(10, l) - n << endl;
      } else {
        int j = l - 1;
        for (int p = 0; p < i; p++) {
          n = n - v[p] * pp(10, j);
          j--;
        }
        cout << pp(10, l - i) - n << endl;
      }
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int pp(int x, int y) {
  long long int ans = 1;
  for (int i = 1; i <= y; i++) {
    ans = ans * x;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    int s;
    cin >> n >> s;
    vector<int> v;
    int sum = 0;
    long long int h = n;
    while (h / 10 != 0) {
      int x = h % 10;
      v.push_back(x);
      sum += x;
      h = h / 10;
    }
    int x = h % 10;
    v.push_back(x);
    sum += x;
    if (sum <= s) {
      cout << "0" << endl;
    } else {
      reverse(v.begin(), v.end());
      int i = 0;
      int ss = 0;
      int l = v.size();
      while (i < l) {
        ss += v[i];
        if (ss < s)
          i++;
        else
          break;
      }
      if (i == 0) {
        cout << pp(10, l) - n << endl;
      } else {
        int j = l - 1;
        for (int p = 0; p < i; p++) {
          n = n - v[p] * pp(10, j);
          j--;
        }
        cout << pp(10, l - i) - n << endl;
      }
    }
  }
  return 0;
}

```