#include <bits/stdc++.h>
using namespace std;
int sum(long long n) {
  int res = 0;
  while (n > 0) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
void solve() {
  long long n, s;
  cin >> n >> s;
  if (sum(n) <= s) {
    cout << "0" << endl;
    return;
  }
  long long ans = 0;
  long long pw = 1;
  for (int i = 0; i < 18; i++) {
    int digit = (n / pw) % 10;
    long long add = pw * ((10 - digit) % 10);
    n += add;
    ans += add;
    if (sum(n) <= s) {
      break;
    }
    pw *= 10;
  }
  cout << ans << endl;
}
int main() {
  long long test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  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;
int sum(long long n) {
  int res = 0;
  while (n > 0) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
void solve() {
  long long n, s;
  cin >> n >> s;
  if (sum(n) <= s) {
    cout << "0" << endl;
    return;
  }
  long long ans = 0;
  long long pw = 1;
  for (int i = 0; i < 18; i++) {
    int digit = (n / pw) % 10;
    long long add = pw * ((10 - digit) % 10);
    n += add;
    ans += add;
    if (sum(n) <= s) {
      break;
    }
    pw *= 10;
  }
  cout << ans << endl;
}
int main() {
  long long test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}

```