#include <bits/stdc++.h>
using namespace std;
long long sum_of_digits(long long n) {
  long long res = 0;
  while (n) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long s;
  cin >> s;
  long long moves = 0;
  int exp = 1;
  while (sum_of_digits(n) > s) {
    long long power_10 = pow(10, exp);
    long long moves_taken = power_10 - n % power_10;
    moves += moves_taken;
    n += moves_taken;
    exp++;
  }
  cout << moves << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  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 sum_of_digits(long long n) {
  long long res = 0;
  while (n) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long s;
  cin >> s;
  long long moves = 0;
  int exp = 1;
  while (sum_of_digits(n) > s) {
    long long power_10 = pow(10, exp);
    long long moves_taken = power_10 - n % power_10;
    moves += moves_taken;
    n += moves_taken;
    exp++;
  }
  cout << moves << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```