#include <bits/stdc++.h>
using namespace std;
long long cif(long long m) { return m % 10; }
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v;
    v.push_back(cif(m));
    long long x = m * 2;
    while (cif(x) != v[0]) {
      v.push_back(cif(x));
      x += m;
    }
    long long sum = 0;
    for (long long i = 0; i < v.size(); i++) sum += v[i];
    x = n / m;
    long long ans = 0;
    ans += (x / v.size()) * sum;
    for (long long i = 0; i < x % v.size(); i++) ans += v[i];
    cout << ans << endl;
  }
}

### 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;
long long cif(long long m) { return m % 10; }
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v;
    v.push_back(cif(m));
    long long x = m * 2;
    while (cif(x) != v[0]) {
      v.push_back(cif(x));
      x += m;
    }
    long long sum = 0;
    for (long long i = 0; i < v.size(); i++) sum += v[i];
    x = n / m;
    long long ans = 0;
    ans += (x / v.size()) * sum;
    for (long long i = 0; i < x % v.size(); i++) ans += v[i];
    cout << ans << endl;
  }
}

```