#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
long long int solve(long long int n) {
  long long int add = 0;
  while (n) {
    add += n % 10;
    n /= 10;
  }
  return add;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s;
    cin >> n >> s;
    long long int power[20];
    power[0] = 1;
    if (solve(n) <= s) {
      cout << 0 << '\n';
      continue;
    }
    for (long long int i = 1; i <= 18; i++) {
      power[i] = power[i - 1] * 10;
    }
    long long int ans = 0;
    for (long long int i = 0; i <= 18; i++) {
      long long int temp = (n / power[i]) % 10;
      long long int toadd = power[i] * ((10 - temp));
      n += toadd;
      ans += toadd;
      if (solve(n) <= s) {
        break;
      }
    }
    cout << ans << '\n';
  }
  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;
const long long int mod = 1e9 + 7;
long long int solve(long long int n) {
  long long int add = 0;
  while (n) {
    add += n % 10;
    n /= 10;
  }
  return add;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s;
    cin >> n >> s;
    long long int power[20];
    power[0] = 1;
    if (solve(n) <= s) {
      cout << 0 << '\n';
      continue;
    }
    for (long long int i = 1; i <= 18; i++) {
      power[i] = power[i - 1] * 10;
    }
    long long int ans = 0;
    for (long long int i = 0; i <= 18; i++) {
      long long int temp = (n / power[i]) % 10;
      long long int toadd = power[i] * ((10 - temp));
      n += toadd;
      ans += toadd;
      if (solve(n) <= s) {
        break;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```