#include <bits/stdc++.h>
using namespace std;
int get_sum(long long x) {
  int sum = 0;
  while (x > 0) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
void solve() {
  long long n;
  int s;
  scanf("%lld%d", &n, &s);
  int sum = get_sum(n);
  long long moves = 0;
  int dg = 0;
  while (sum > s) {
    long long tmp = n;
    for (int i = 0; i < dg; ++i) tmp /= 10;
    if (tmp % 10 == 0) {
      ++dg;
      continue;
    }
    long long increment = (long long)pow(10, dg) * (10 - (tmp % 10));
    moves += increment;
    n += increment;
    sum = get_sum(n);
    ++dg;
  }
  printf("%lld\n", moves);
}
int main() {
  int TC;
  scanf("%d", &TC);
  while (TC--) solve();
  return 0;
}

### Prompt

Please formulate 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;
int get_sum(long long x) {
  int sum = 0;
  while (x > 0) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
void solve() {
  long long n;
  int s;
  scanf("%lld%d", &n, &s);
  int sum = get_sum(n);
  long long moves = 0;
  int dg = 0;
  while (sum > s) {
    long long tmp = n;
    for (int i = 0; i < dg; ++i) tmp /= 10;
    if (tmp % 10 == 0) {
      ++dg;
      continue;
    }
    long long increment = (long long)pow(10, dg) * (10 - (tmp % 10));
    moves += increment;
    n += increment;
    sum = get_sum(n);
    ++dg;
  }
  printf("%lld\n", moves);
}
int main() {
  int TC;
  scanf("%d", &TC);
  while (TC--) solve();
  return 0;
}

```