#include <bits/stdc++.h>
using namespace std;
void test_case() {
  long long int n, s;
  cin >> n >> s;
  vector<long long int> a(19, 0);
  vector<long long int> b(19, 0);
  long long int t = n;
  for (int i = 0; t > 0; i++) {
    a[i] = t % 10;
    t /= 10;
  }
  long long int sum = accumulate(a.begin(), a.end(), 0LL);
  for (int i = 0; sum > s; i++) {
    if (a[i] > 0) {
      sum -= (a[i] - 1);
      b[i] = 10 - a[i];
      a[i + 1]++;
    }
  }
  long long int ans = 0;
  long long int j = 1;
  for (int i = 0; i < 18; i++, j *= 10) {
    ans += b[i] * j;
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tc = 1;
  cin >> tc;
  while (tc--) test_case();
  return 0;
}

### Prompt

Your challenge is to write 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;
void test_case() {
  long long int n, s;
  cin >> n >> s;
  vector<long long int> a(19, 0);
  vector<long long int> b(19, 0);
  long long int t = n;
  for (int i = 0; t > 0; i++) {
    a[i] = t % 10;
    t /= 10;
  }
  long long int sum = accumulate(a.begin(), a.end(), 0LL);
  for (int i = 0; sum > s; i++) {
    if (a[i] > 0) {
      sum -= (a[i] - 1);
      b[i] = 10 - a[i];
      a[i + 1]++;
    }
  }
  long long int ans = 0;
  long long int j = 1;
  for (int i = 0; i < 18; i++, j *= 10) {
    ans += b[i] * j;
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tc = 1;
  cin >> tc;
  while (tc--) test_case();
  return 0;
}

```