#include <bits/stdc++.h>
int tcase, m;
long long n;
int gao(long long x) {
  int ret = 0;
  for (; x; x /= 10) ret += x % 10;
  return ret;
}
int main() {
  scanf("%d", &tcase);
  while (tcase--) {
    scanf("%lld%d", &n, &m);
    if (gao(n) <= m) {
      printf("0\n");
      continue;
    }
    long long x = 10;
    while (gao(n + (x - n % x)) > m) x *= 10;
    printf("%lld\n", x - n % x);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
int tcase, m;
long long n;
int gao(long long x) {
  int ret = 0;
  for (; x; x /= 10) ret += x % 10;
  return ret;
}
int main() {
  scanf("%d", &tcase);
  while (tcase--) {
    scanf("%lld%d", &n, &m);
    if (gao(n) <= m) {
      printf("0\n");
      continue;
    }
    long long x = 10;
    while (gao(n + (x - n % x)) > m) x *= 10;
    printf("%lld\n", x - n % x);
  }
  return 0;
}

```