#include <bits/stdc++.h>
using namespace std;
long long a[150500];
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  vector<long long> ans;
  for (int f = 0; f < t; f++) {
    long long n, m;
    cin >> n >> m;
    vector<long long> l;
    l.push_back(m % 10);
    for (long long i = m + m; l[0] != i % 10; i += m) l.push_back(i % 10);
    long long c = n / m / l.size() * accumulate(l.begin(), l.end(), 0);
    for (int i = 0; i < n / m % l.size(); i++) c += l[i];
    ans.push_back(c);
  }
  for (auto i : ans) cout << i << endl;
  return 0;
}

### Prompt

Generate 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;
long long a[150500];
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  vector<long long> ans;
  for (int f = 0; f < t; f++) {
    long long n, m;
    cin >> n >> m;
    vector<long long> l;
    l.push_back(m % 10);
    for (long long i = m + m; l[0] != i % 10; i += m) l.push_back(i % 10);
    long long c = n / m / l.size() * accumulate(l.begin(), l.end(), 0);
    for (int i = 0; i < n / m % l.size(); i++) c += l[i];
    ans.push_back(c);
  }
  for (auto i : ans) cout << i << endl;
  return 0;
}

```