#include <bits/stdc++.h>
using namespace std;
const int NUM = 2e5 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int num_tests = 1;
  cin >> num_tests;
  while (num_tests-- > 0) {
    long long int n, m;
    cin >> n >> m;
    long long int div = n / m;
    long long int sum[10];
    sum[0] = 0;
    for (int i = 1; i < 10; i++) sum[i] = sum[i - 1] + (i * m) % 10;
    long long int ans = (div / 10) * sum[9];
    ans += sum[(n / m) % 10];
    cout << ans << "\n";
  }
}

### Prompt

Your task is to create 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;
const int NUM = 2e5 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int num_tests = 1;
  cin >> num_tests;
  while (num_tests-- > 0) {
    long long int n, m;
    cin >> n >> m;
    long long int div = n / m;
    long long int sum[10];
    sum[0] = 0;
    for (int i = 1; i < 10; i++) sum[i] = sum[i - 1] + (i * m) % 10;
    long long int ans = (div / 10) * sum[9];
    ans += sum[(n / m) % 10];
    cout << ans << "\n";
  }
}

```