#include <bits/stdc++.h>
using namespace std;
int q;
long long n, m;
int vis[100];
int main() {
  cin >> q;
  while (q--) {
    cin >> n >> m;
    long long stp = n / m;
    long long lst = m % 10;
    long long sum = 0;
    memset(vis, false, sizeof(vis));
    int d = 0;
    for (int i = 1; i <= 10; i++) {
      if (vis[lst * i % 10]) break;
      d++;
      vis[lst * i % 10] = true;
      sum += lst * i % 10;
    }
    long long ans = sum * (stp / d);
    long long lft = stp % d;
    for (int i = 1; i <= lft; i++) {
      ans += lst * i % 10;
    }
    cout << ans << 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 q;
long long n, m;
int vis[100];
int main() {
  cin >> q;
  while (q--) {
    cin >> n >> m;
    long long stp = n / m;
    long long lst = m % 10;
    long long sum = 0;
    memset(vis, false, sizeof(vis));
    int d = 0;
    for (int i = 1; i <= 10; i++) {
      if (vis[lst * i % 10]) break;
      d++;
      vis[lst * i % 10] = true;
      sum += lst * i % 10;
    }
    long long ans = sum * (stp / d);
    long long lft = stp % d;
    for (int i = 1; i <= lft; i++) {
      ans += lst * i % 10;
    }
    cout << ans << endl;
  }
}

```