#include <bits/stdc++.h>
using namespace std;
long long sum_of_digits(long long n) {
  int r, s = 0;
  while (n) {
    r = n % 10;
    s += r;
    n = n / 10;
  }
  return s;
}
long long convert(long long n) {
  int p = 0, r = 0;
  long long new_num, k = n;
  while (r == 0) {
    p++;
    r = n % 10;
    n = n / 10;
  }
  new_num = k + pow(10, p) - (r * pow(10, p - 1));
  return new_num;
}
int main() {
  int t;
  cin >> t;
  for (int _ = 0; _ < t; _++) {
    long long n, s, temp;
    cin >> n >> s;
    temp = n;
    while (sum_of_digits(temp) > s) {
      temp = convert(temp);
    }
    cout << temp - n << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long sum_of_digits(long long n) {
  int r, s = 0;
  while (n) {
    r = n % 10;
    s += r;
    n = n / 10;
  }
  return s;
}
long long convert(long long n) {
  int p = 0, r = 0;
  long long new_num, k = n;
  while (r == 0) {
    p++;
    r = n % 10;
    n = n / 10;
  }
  new_num = k + pow(10, p) - (r * pow(10, p - 1));
  return new_num;
}
int main() {
  int t;
  cin >> t;
  for (int _ = 0; _ < t; _++) {
    long long n, s, temp;
    cin >> n >> s;
    temp = n;
    while (sum_of_digits(temp) > s) {
      temp = convert(temp);
    }
    cout << temp - n << "\n";
  }
  return 0;
}

```