#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    long long n, m;
    cin >> n >> m;
    if (n < m) {
      cout << 0 << '\n';
      continue;
    }
    vector<long long> a(10);
    long long val = m, sum = 0;
    for (long long i = 0; i < 10; ++i) {
      a[i] = val % 10;
      val += m;
      sum += a[i];
    }
    long long l = n / m;
    long long x = l % 10;
    long long y = l / 10;
    long long res = sum * y;
    for (int i = 0; i < x; ++i) res += a[i];
    cout << res << '\n';
  }
  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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    long long n, m;
    cin >> n >> m;
    if (n < m) {
      cout << 0 << '\n';
      continue;
    }
    vector<long long> a(10);
    long long val = m, sum = 0;
    for (long long i = 0; i < 10; ++i) {
      a[i] = val % 10;
      val += m;
      sum += a[i];
    }
    long long l = n / m;
    long long x = l % 10;
    long long y = l / 10;
    long long res = sum * y;
    for (int i = 0; i < x; ++i) res += a[i];
    cout << res << '\n';
  }
  return 0;
}

```