#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, m;
    cin >> n >> m;
    int ans[11][11];
    for (long long i = 0; i < (11); ++i)
      for (long long j = 0; j < (11); ++j) ans[i][j] = 0;
    for (int i = 1; i < 11; i++) {
      for (int j = 1; j < 11; j++) {
        ans[i][j] = ans[i][j - 1] + (i * j) % 10;
      }
    }
    cout << ((n / m - (n / m) % 10) / 10) * ans[m % 10][10] +
                ans[m % 10][((n / m) % 10)]
         << "\n";
  }
}

### Prompt

Your challenge is to write 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() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, m;
    cin >> n >> m;
    int ans[11][11];
    for (long long i = 0; i < (11); ++i)
      for (long long j = 0; j < (11); ++j) ans[i][j] = 0;
    for (int i = 1; i < 11; i++) {
      for (int j = 1; j < 11; j++) {
        ans[i][j] = ans[i][j - 1] + (i * j) % 10;
      }
    }
    cout << ((n / m - (n / m) % 10) / 10) * ans[m % 10][10] +
                ans[m % 10][((n / m) % 10)]
         << "\n";
  }
}

```