#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v(0);
    long long temp = m % 10, next = m % 10, last_dig = m % 10, sum = 0,
              count = 0;
    while (last_dig != 0) {
      last_dig = next % 10;
      v.push_back(last_dig);
      sum += last_dig;
      count++;
      next += temp;
    }
    long long q = (n / m);
    if (q == 0 || count == 0) {
      cout << 0 << endl;
    } else {
      long long sum1 = 0;
      for (int i = 0; i < q % count; i++) {
        sum1 += v[i];
      }
      long long ans = (q / count) * sum + sum1;
      cout << ans << endl;
    }
    v.clear();
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v(0);
    long long temp = m % 10, next = m % 10, last_dig = m % 10, sum = 0,
              count = 0;
    while (last_dig != 0) {
      last_dig = next % 10;
      v.push_back(last_dig);
      sum += last_dig;
      count++;
      next += temp;
    }
    long long q = (n / m);
    if (q == 0 || count == 0) {
      cout << 0 << endl;
    } else {
      long long sum1 = 0;
      for (int i = 0; i < q % count; i++) {
        sum1 += v[i];
      }
      long long ans = (q / count) * sum + sum1;
      cout << ans << endl;
    }
    v.clear();
  }
}

```