#include <bits/stdc++.h>
using namespace std;
int main(void) {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    long long n, m;
    scanf("%lld %lld", &n, &m);
    int last_num = m % 10;
    long long div = n / m;
    vector<int> v;
    int cycle = 0;
    int sum = last_num;
    int cnt = 0;
    while (sum != 0) {
      cycle += sum;
      sum += last_num;
      sum %= 10;
      cnt++;
      v.push_back(cycle);
    }
    v.push_back(cycle + 0);
    cnt++;
    long long ddiv = div / cnt;
    long long remain = div % cnt;
    long long result = ddiv * v[cnt - 1];
    if (remain != 0) result += v[remain - 1];
    printf("%lld\n", result);
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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(void) {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    long long n, m;
    scanf("%lld %lld", &n, &m);
    int last_num = m % 10;
    long long div = n / m;
    vector<int> v;
    int cycle = 0;
    int sum = last_num;
    int cnt = 0;
    while (sum != 0) {
      cycle += sum;
      sum += last_num;
      sum %= 10;
      cnt++;
      v.push_back(cycle);
    }
    v.push_back(cycle + 0);
    cnt++;
    long long ddiv = div / cnt;
    long long remain = div % cnt;
    long long result = ddiv * v[cnt - 1];
    if (remain != 0) result += v[remain - 1];
    printf("%lld\n", result);
  }
  return 0;
}

```