#include <bits/stdc++.h>
using namespace std;
long long int give_sum(long long int n) {
  long long int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long int solve(long long int n, long long int s) {
  if (give_sum(n) <= s) return 0;
  long long int go = 0;
  while (n % 10) {
    go++, n++;
  }
  return go + solve(n / 10, s) * 10;
}
void solve() {
  long long int s, i, j, y, pw = 1, x, z, moves = 0, prev = 0, sum = 0, cnt = 0,
                            n;
  cin >> n >> s;
  cout << solve(n, s) << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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 int give_sum(long long int n) {
  long long int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long int solve(long long int n, long long int s) {
  if (give_sum(n) <= s) return 0;
  long long int go = 0;
  while (n % 10) {
    go++, n++;
  }
  return go + solve(n / 10, s) * 10;
}
void solve() {
  long long int s, i, j, y, pw = 1, x, z, moves = 0, prev = 0, sum = 0, cnt = 0,
                            n;
  cin >> n >> s;
  cout << solve(n, s) << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```