#include <bits/stdc++.h>
using namespace std;
int main() {
  long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long last_digit = m % 10;
    if (last_digit == 0)
      cout << 0 << endl;
    else {
      long rem[10];
      rem[0] = 0;
      for (long i = 1; i <= 9; i++) {
        rem[i] = (last_digit * i) % 10;
      }
      long long sum[10];
      sum[0] = 0;
      for (long long i = 1; i <= 9; i++) sum[i] = sum[i - 1] + rem[i];
      long long total_rounds = n / (m * 10);
      long long count = total_rounds * sum[9];
      long long remaining = (n % (m * 10)) / m;
      count = count + sum[remaining];
      cout << count << endl;
    }
  }
}

### Prompt

Generate a CPP solution to the following problem:

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() {
  long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long last_digit = m % 10;
    if (last_digit == 0)
      cout << 0 << endl;
    else {
      long rem[10];
      rem[0] = 0;
      for (long i = 1; i <= 9; i++) {
        rem[i] = (last_digit * i) % 10;
      }
      long long sum[10];
      sum[0] = 0;
      for (long long i = 1; i <= 9; i++) sum[i] = sum[i - 1] + rem[i];
      long long total_rounds = n / (m * 10);
      long long count = total_rounds * sum[9];
      long long remaining = (n % (m * 10)) / m;
      count = count + sum[remaining];
      cout << count << endl;
    }
  }
}

```