#include <bits/stdc++.h>
using namespace std;
int sumdigit(long long int a, vector<int> &f) {
  while (a > 0) {
    f.push_back(a % 10);
    a = a / 10;
  }
  int ans = 0;
  reverse(f.begin(), f.end());
  for (int i = 0; i < f.size(); i++) {
    ans += f[i];
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t > 0) {
    long long int a, b;
    cin >> a >> b;
    long long int c = 0;
    vector<int> f;
    int k = sumdigit(a, f);
    if (k <= b) {
      c = 0;
    } else {
      int x = 0;
      int i = 0;
      for (i = 0; i < f.size(); i++) {
        x += f[i];
        if (x >= b) {
          break;
        }
      }
      if (i == 0) {
        long long int kl = pow(10, f.size());
        c = kl - a;
      } else {
        long long int y = pow(10, f.size() - i);
        long long int v = 0;
        long long int p = 1;
        for (int j = f.size() - 1; j > i - 1; j--) {
          v += f[j] * p;
          p = p * 10;
        }
        c = y - v;
      }
    }
    cout << c << endl;
    t--;
  }
}

### 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 sumdigit(long long int a, vector<int> &f) {
  while (a > 0) {
    f.push_back(a % 10);
    a = a / 10;
  }
  int ans = 0;
  reverse(f.begin(), f.end());
  for (int i = 0; i < f.size(); i++) {
    ans += f[i];
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t > 0) {
    long long int a, b;
    cin >> a >> b;
    long long int c = 0;
    vector<int> f;
    int k = sumdigit(a, f);
    if (k <= b) {
      c = 0;
    } else {
      int x = 0;
      int i = 0;
      for (i = 0; i < f.size(); i++) {
        x += f[i];
        if (x >= b) {
          break;
        }
      }
      if (i == 0) {
        long long int kl = pow(10, f.size());
        c = kl - a;
      } else {
        long long int y = pow(10, f.size() - i);
        long long int v = 0;
        long long int p = 1;
        for (int j = f.size() - 1; j > i - 1; j--) {
          v += f[j] * p;
          p = p * 10;
        }
        c = y - v;
      }
    }
    cout << c << endl;
    t--;
  }
}

```