#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  vector<int> a{1, 10, 5, 10, 5, 2, 5, 10, 5, 10};
  int sum[] = {0, 45, 20, 45, 20, 5, 20, 45, 20, 45};
  while (t--) {
    long long int n, m;
    cin >> n >> m;
    if (m > n) {
      cout << "0" << endl;
      continue;
    }
    long long x = n / m;
    long long int z = m;
    int last = m % 10;
    long long val = 0;
    long long y = x / a[last];
    long long rem = x % a[last];
    val += y * sum[last];
    while (rem > 0) {
      val += (z % 10);
      z += m;
      rem--;
    }
    cout << val << endl;
  }
}

### 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;
  vector<int> a{1, 10, 5, 10, 5, 2, 5, 10, 5, 10};
  int sum[] = {0, 45, 20, 45, 20, 5, 20, 45, 20, 45};
  while (t--) {
    long long int n, m;
    cin >> n >> m;
    if (m > n) {
      cout << "0" << endl;
      continue;
    }
    long long x = n / m;
    long long int z = m;
    int last = m % 10;
    long long val = 0;
    long long y = x / a[last];
    long long rem = x % a[last];
    val += y * sum[last];
    while (rem > 0) {
      val += (z % 10);
      z += m;
      rem--;
    }
    cout << val << endl;
  }
}

```