#include <bits/stdc++.h>
using namespace std;
inline long long rd() {
  long long x;
  cin >> x;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long T = 1;
  cin >> T;
  while (T--) {
    long long n, s;
    cin >> n >> s;
    long long sum = 0;
    long long nn = n;
    while (nn > 0) {
      sum += nn % 10;
      nn /= 10;
    }
    if (sum <= s) {
      cout << "0\n";
      continue;
    }
    nn = n;
    long long mul = 1;
    long long ans = 0;
    while (sum > s) {
      long long num = (nn / mul) % 10;
      if (num != 0) {
        ans += (10 - num) * mul;
        nn += (10 - num) * mul;
      }
      sum = 0;
      long long nnn = nn;
      while (nnn > 0) {
        sum += nnn % 10;
        nnn /= 10;
      }
      mul *= 10;
    }
    cout << ans << "\n";
  }
  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;
inline long long rd() {
  long long x;
  cin >> x;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long T = 1;
  cin >> T;
  while (T--) {
    long long n, s;
    cin >> n >> s;
    long long sum = 0;
    long long nn = n;
    while (nn > 0) {
      sum += nn % 10;
      nn /= 10;
    }
    if (sum <= s) {
      cout << "0\n";
      continue;
    }
    nn = n;
    long long mul = 1;
    long long ans = 0;
    while (sum > s) {
      long long num = (nn / mul) % 10;
      if (num != 0) {
        ans += (10 - num) * mul;
        nn += (10 - num) * mul;
      }
      sum = 0;
      long long nnn = nn;
      while (nnn > 0) {
        sum += nnn % 10;
        nnn /= 10;
      }
      mul *= 10;
    }
    cout << ans << "\n";
  }
  return 0;
}

```