#include <bits/stdc++.h>
int main(void) {
  int t, s, sum;
  uint64_t n, moves, copy;
  scanf("%i", &t);
  for (; t; t--) {
    scanf("%llu %i", &n, &s);
    moves = 0;
    sum = 0;
    copy = n;
    while (copy) {
      sum += copy % 10;
      copy /= 10;
    }
    copy = n;
    if (sum <= s) {
      printf("%llu\n", moves);
      continue;
    }
    sum++;
    int i;
    for (i = 0; sum > s; i++) {
      sum -= copy % 10;
      copy /= 10;
    }
    copy++;
    for (; i; --i) {
      copy *= 10;
    }
    moves = copy - n;
    printf("%llu\n", moves);
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 main(void) {
  int t, s, sum;
  uint64_t n, moves, copy;
  scanf("%i", &t);
  for (; t; t--) {
    scanf("%llu %i", &n, &s);
    moves = 0;
    sum = 0;
    copy = n;
    while (copy) {
      sum += copy % 10;
      copy /= 10;
    }
    copy = n;
    if (sum <= s) {
      printf("%llu\n", moves);
      continue;
    }
    sum++;
    int i;
    for (i = 0; sum > s; i++) {
      sum -= copy % 10;
      copy /= 10;
    }
    copy++;
    for (; i; --i) {
      copy *= 10;
    }
    moves = copy - n;
    printf("%llu\n", moves);
  }
}

```