#include <bits/stdc++.h>
using namespace std;
const long long int M = 1e9 + 7;
long long int d_sum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
void solve() {
  long long int n, s;
  cin >> n >> s;
  long long int val = n;
  if (d_sum(n) <= s) {
    cout << 0 << endl;
    return;
  }
  int dig = 1;
  while (1) {
    for (int i = 0; i < dig; i++) val = val / 10;
    val++;
    for (int i = 0; i < dig; i++) val = val * 10;
    dig++;
    if (d_sum(val) <= s) {
      cout << val - n << endl;
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please formulate 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;
const long long int M = 1e9 + 7;
long long int d_sum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
void solve() {
  long long int n, s;
  cin >> n >> s;
  long long int val = n;
  if (d_sum(n) <= s) {
    cout << 0 << endl;
    return;
  }
  int dig = 1;
  while (1) {
    for (int i = 0; i < dig; i++) val = val / 10;
    val++;
    for (int i = 0; i < dig; i++) val = val * 10;
    dig++;
    if (d_sum(val) <= s) {
      cout << val - n << endl;
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) solve();
}

```