#include <bits/stdc++.h>
using namespace std;
long long a[150500], l[150500], m = 1e7, mn = 1e7, ans = 0;
int main() {
  int t;
  cin >> t;
  while (t--) {
    ans = 0;
    long long x, y;
    cin >> x >> y;
    l[0] = y % 10ll;
    int i = 1;
    while (1) {
      l[i] = (y * (i + 1ll)) % 10ll;
      if (l[i] == l[0]) break;
      i++;
    }
    for (int j = 0; j < i; j++) {
      ans += l[j];
    }
    ans *= x / (y * (i));
    for (int j = 0; j < (x % (y * (i))) / y; j++) {
      ans += l[j];
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
long long a[150500], l[150500], m = 1e7, mn = 1e7, ans = 0;
int main() {
  int t;
  cin >> t;
  while (t--) {
    ans = 0;
    long long x, y;
    cin >> x >> y;
    l[0] = y % 10ll;
    int i = 1;
    while (1) {
      l[i] = (y * (i + 1ll)) % 10ll;
      if (l[i] == l[0]) break;
      i++;
    }
    for (int j = 0; j < i; j++) {
      ans += l[j];
    }
    ans *= x / (y * (i));
    for (int j = 0; j < (x % (y * (i))) / y; j++) {
      ans += l[j];
    }
    cout << ans << endl;
  }
  return 0;
}

```