#include <bits/stdc++.h>
using namespace std;
void print() { cout << endl; }
template <typename First, typename... Strings>
void print(First arg, const Strings&... rest) {
  cout << arg << " ";
  print(rest...);
}
void test_case() {
  long long int n, s;
  cin >> n >> s;
  long long int sum = 0;
  vector<long long int> d;
  while (n > 0) {
    long long int r = n % 10;
    sum += r;
    d.push_back(r);
    n /= 10;
  }
  long long int j = 0, cnt = 0, mul = 1;
  while (sum > s) {
    if (d[j] != 0) {
      cnt += (10 - d[j]) * mul;
      sum -= d[j] - 1;
      long long int k = j + 1;
      while (k < d.size() && d[k] == 9) {
        sum -= 9;
        d[k] = 0;
        k++;
      }
      if (k < d.size()) d[k]++;
    }
    j++;
    mul *= 10;
  }
  print(cnt);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    test_case();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 print() { cout << endl; }
template <typename First, typename... Strings>
void print(First arg, const Strings&... rest) {
  cout << arg << " ";
  print(rest...);
}
void test_case() {
  long long int n, s;
  cin >> n >> s;
  long long int sum = 0;
  vector<long long int> d;
  while (n > 0) {
    long long int r = n % 10;
    sum += r;
    d.push_back(r);
    n /= 10;
  }
  long long int j = 0, cnt = 0, mul = 1;
  while (sum > s) {
    if (d[j] != 0) {
      cnt += (10 - d[j]) * mul;
      sum -= d[j] - 1;
      long long int k = j + 1;
      while (k < d.size() && d[k] == 9) {
        sum -= 9;
        d[k] = 0;
        k++;
      }
      if (k < d.size()) d[k]++;
    }
    j++;
    mul *= 10;
  }
  print(cnt);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    test_case();
  }
  return 0;
}

```