#include <bits/stdc++.h>
using namespace std;
long long getnum(long long x) {
  long long ans = 1;
  while (x) {
    ans *= 10;
    x /= 10;
  }
  return ans / 10;
}
long long getans(long long a, long long b, int k) {
  while (k - a / b > 0) {
    k -= a / b;
    a %= b;
    b /= 10;
  }
  return a;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    scanf("%lld %lld", &n, &s);
    long long tn = n;
    long long tmp3 = 0;
    while (tn) {
      tmp3 += tn % 10;
      tn /= 10;
    }
    if (tmp3 <= s)
      printf("0\n");
    else {
      long long tmp1 = getnum(n);
      long long tmp2 = getans(n, tmp1, s);
      long long ans = getnum(tmp2) * 10 - tmp2;
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 getnum(long long x) {
  long long ans = 1;
  while (x) {
    ans *= 10;
    x /= 10;
  }
  return ans / 10;
}
long long getans(long long a, long long b, int k) {
  while (k - a / b > 0) {
    k -= a / b;
    a %= b;
    b /= 10;
  }
  return a;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    scanf("%lld %lld", &n, &s);
    long long tn = n;
    long long tmp3 = 0;
    while (tn) {
      tmp3 += tn % 10;
      tn /= 10;
    }
    if (tmp3 <= s)
      printf("0\n");
    else {
      long long tmp1 = getnum(n);
      long long tmp2 = getans(n, tmp1, s);
      long long ans = getnum(tmp2) * 10 - tmp2;
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```