#include <bits/stdc++.h>
using namespace std;
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    long long n, m;
    cin >> n >> m;
    long long ans = 0;
    long long ten = 0;
    if (m * 10 >= n) {
      int ii = 1;
      for (long long i = m; i <= n; i = m * ++ii) {
        ans += (i % 10);
      }
      cout << ans << endl;
      continue;
    }
    for (int i = 1; i <= 10; i++) {
      int x_1 = (m * i) % 10;
      ten += x_1;
    }
    long long n1 = n / (m * 10);
    ans = n1 * ten;
    if (n / m == 0) continue;
    for (long long i = n1 * (m * 10); i <= n; i += m) {
      ans += (i % 10);
    }
    cout << ans << endl;
  }
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    long long n, m;
    cin >> n >> m;
    long long ans = 0;
    long long ten = 0;
    if (m * 10 >= n) {
      int ii = 1;
      for (long long i = m; i <= n; i = m * ++ii) {
        ans += (i % 10);
      }
      cout << ans << endl;
      continue;
    }
    for (int i = 1; i <= 10; i++) {
      int x_1 = (m * i) % 10;
      ten += x_1;
    }
    long long n1 = n / (m * 10);
    ans = n1 * ten;
    if (n / m == 0) continue;
    for (long long i = n1 * (m * 10); i <= n; i += m) {
      ans += (i % 10);
    }
    cout << ans << endl;
  }
}

```