#include <bits/stdc++.h>
using namespace std;
const long long Max = 1e6;
long long n;
long long sum(long long x) {
  long long sum = 0;
  while (x) {
    sum += (x % 10);
    x /= 10;
  }
  return sum;
}
long long ar[Max];
void normtest() {
  long long t;
  cin >> t;
  while (t--) {
    long long m, k, pro = 1, cnt = 0, x, y, z;
    bool flag = 0;
    cin >> n;
    cin >> k;
    x = 10;
    do {
      y = n;
      long long va = sum(y);
      if (va <= k) {
        break;
      }
      n = y / x * x + x;
      x *= 10;
      cnt += n - y;
    } while (1);
    cout << cnt;
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  normtest();
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const long long Max = 1e6;
long long n;
long long sum(long long x) {
  long long sum = 0;
  while (x) {
    sum += (x % 10);
    x /= 10;
  }
  return sum;
}
long long ar[Max];
void normtest() {
  long long t;
  cin >> t;
  while (t--) {
    long long m, k, pro = 1, cnt = 0, x, y, z;
    bool flag = 0;
    cin >> n;
    cin >> k;
    x = 10;
    do {
      y = n;
      long long va = sum(y);
      if (va <= k) {
        break;
      }
      n = y / x * x + x;
      x *= 10;
      cnt += n - y;
    } while (1);
    cout << cnt;
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  normtest();
  return 0;
}

```