#include <bits/stdc++.h>
using namespace std;
long long int ws(long long int i) {
  int ss = 0;
  while (i) {
    ss += i % 10;
    i /= 10;
  }
  return ss;
}
int main() {
  long long int t, n, s, k, b, nn;
  ;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld", &n, &s);
    nn = n;
    k = ws(n);
    b = 10;
    while (k > s) {
      long long int d = n % b;
      d = b - d;
      n += d;
      b *= 10;
      k = ws(n);
    }
    printf("%lld\n", n - nn);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int ws(long long int i) {
  int ss = 0;
  while (i) {
    ss += i % 10;
    i /= 10;
  }
  return ss;
}
int main() {
  long long int t, n, s, k, b, nn;
  ;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld", &n, &s);
    nn = n;
    k = ws(n);
    b = 10;
    while (k > s) {
      long long int d = n % b;
      d = b - d;
      n += d;
      b *= 10;
      k = ws(n);
    }
    printf("%lld\n", n - nn);
  }
  return 0;
}

```