#include <bits/stdc++.h>
using namespace std;
long long getDigitSum(long long x) {
  long long sum = 0;
  while (x > 0) {
    sum += (x % 10);
    x /= 10;
  }
  return sum;
}
long long getMovesCount(long long n, long long s) {
  if (getDigitSum(n) <= s) return 0;
  long long pw = 1;
  long long res = 0;
  for (long long i = 0; i < 18; i++) {
    long long digi = (n / pw) % 10;
    long long mv = ((10 - digi) % 10) * pw;
    n += mv;
    res += mv;
    if (getDigitSum(n) <= s) break;
    pw *= 10;
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long s, n;
    cin >> n >> s;
    cout << getMovesCount(n, s) << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long getDigitSum(long long x) {
  long long sum = 0;
  while (x > 0) {
    sum += (x % 10);
    x /= 10;
  }
  return sum;
}
long long getMovesCount(long long n, long long s) {
  if (getDigitSum(n) <= s) return 0;
  long long pw = 1;
  long long res = 0;
  for (long long i = 0; i < 18; i++) {
    long long digi = (n / pw) % 10;
    long long mv = ((10 - digi) % 10) * pw;
    n += mv;
    res += mv;
    if (getDigitSum(n) <= s) break;
    pw *= 10;
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long s, n;
    cin >> n >> s;
    cout << getMovesCount(n, s) << endl;
  }
  return 0;
}

```