#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long last = n - (n % m);
    long long count = 1;
    long long sum = 0;
    long long first = n % m;
    long long tmp = 2 * m;
    sum = m % 10;
    while (tmp % 10 != m % 10) {
      sum += tmp % 10;
      tmp = tmp + m;
      count++;
    }
    long long answer = sum * ((n / m) / count);
    long long fac = (n / m) % count;
    tmp = m;
    while (fac--) {
      answer += tmp % 10;
      tmp = tmp + m;
    }
    cout << answer << "\n";
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

Polycarp is reading a book consisting of n pages numbered from 1 to n. Every time he finishes the page with the number divisible by m, he writes down the last digit of this page number. For example, if n=15 and m=5, pages divisible by m are 5, 10, 15. Their last digits are 5, 0, 5 correspondingly, their sum is 10.

Your task is to calculate the sum of all digits Polycarp has written down.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 1000) — the number of queries.

The following q lines contain queries, one per line. Each query is given as two integers n and m (1 ≤ n, m ≤ 10^{16}) — the number of pages in the book and required divisor, respectively.

Output

For each query print the answer for it — the sum of digits written down by Polycarp.

Example

Input


7
1 1
10 1
100 3
1024 14
998244353 1337
123 144
1234312817382646 13


Output


1
45
153
294
3359835
0
427262129093995

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long last = n - (n % m);
    long long count = 1;
    long long sum = 0;
    long long first = n % m;
    long long tmp = 2 * m;
    sum = m % 10;
    while (tmp % 10 != m % 10) {
      sum += tmp % 10;
      tmp = tmp + m;
      count++;
    }
    long long answer = sum * ((n / m) / count);
    long long fac = (n / m) % count;
    tmp = m;
    while (fac--) {
      answer += tmp % 10;
      tmp = tmp + m;
    }
    cout << answer << "\n";
  }
}

```