#include <bits/stdc++.h>
using namespace std;
int tc = 1;
void solve();
int main() {
  scanf("%d", &tc);
  while (tc--) solve();
}
int dig(long long k) {
  int x = 0;
  while (k > 0) {
    x += k % 10;
    k /= 10;
  }
  return x;
}
long long app(long long num, long long pad) {
  for (int i = 0; i < pad; i++) num *= 10;
  return num;
}
void solve() {
  long long n, s, p;
  cin >> n >> s;
  long long m = n;
  long long a = 0;
  if (dig(n) <= s) {
    printf("0\n");
    return;
  }
  while (m >= 0) {
    a++;
    m /= 10;
    p = app(m + 1, a);
    if (dig(p) <= s) {
      printf("%lld\n", p - n);
      return;
    }
  }
}

### 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;
int tc = 1;
void solve();
int main() {
  scanf("%d", &tc);
  while (tc--) solve();
}
int dig(long long k) {
  int x = 0;
  while (k > 0) {
    x += k % 10;
    k /= 10;
  }
  return x;
}
long long app(long long num, long long pad) {
  for (int i = 0; i < pad; i++) num *= 10;
  return num;
}
void solve() {
  long long n, s, p;
  cin >> n >> s;
  long long m = n;
  long long a = 0;
  if (dig(n) <= s) {
    printf("0\n");
    return;
  }
  while (m >= 0) {
    a++;
    m /= 10;
    p = app(m + 1, a);
    if (dig(p) <= s) {
      printf("%lld\n", p - n);
      return;
    }
  }
}

```