#include <bits/stdc++.h>
using namespace std;
set<int> s;
long long a[20];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    s.clear();
    long long n, m;
    long long i = 1;
    for (i = 0; i <= 10; i++) a[i] = 0;
    cin >> n >> m;
    long long co = 0;
    long long sum = 0;
    for (i = 1;; i++) {
      co += m;
      if (s.find(co % 10) != s.end()) break;
      a[co % 10] = co;
      s.insert(co % 10);
      sum += co % 10;
    }
    co -= m;
    long long md = n % co;
    co = n / co;
    long long ans = sum * co;
    for (i = 0; i < 10; i++) {
      if (a[i] <= md)
        if (a[i] % m == 0) ans += a[i] % 10;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
set<int> s;
long long a[20];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    s.clear();
    long long n, m;
    long long i = 1;
    for (i = 0; i <= 10; i++) a[i] = 0;
    cin >> n >> m;
    long long co = 0;
    long long sum = 0;
    for (i = 1;; i++) {
      co += m;
      if (s.find(co % 10) != s.end()) break;
      a[co % 10] = co;
      s.insert(co % 10);
      sum += co % 10;
    }
    co -= m;
    long long md = n % co;
    co = n / co;
    long long ans = sum * co;
    for (i = 0; i < 10; i++) {
      if (a[i] <= md)
        if (a[i] % m == 0) ans += a[i] % 10;
    }
    cout << ans << endl;
  }
  return 0;
}

```