#include <bits/stdc++.h>
using namespace std;
long long ansew(long long n, long long m) {
  long long ex;
  long long rep;
  long long ans = 0;
  ex = n / m;
  int undig = m % 10;
  switch (undig) {
    case 1: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += i;
      }
      return ans;
    }
    case 2: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (2 * i) % 10;
      }
      return ans;
    }
    case 3: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += ((3 * i) % 10);
      }
      return ans;
    }
    case 4: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (4 * i) % 10;
      }
      return ans;
    }
    case 5: {
      rep = ex / 2;
      ans += (5 * rep);
      for (int i = 1; i <= ex % 2; i++) {
        ans += (5 * i) % 10;
      }
      return ans;
    }
    case 6: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (6 * i) % 10;
      }
      return ans;
    }
    case 7: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += ((7 * i) % 10);
      }
      return ans;
    }
    case 8: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (8 * i) % 10;
      }
      return ans;
    }
    case 9: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += (9 * i) % 10;
      }
      return ans;
    }
    case 0: {
      ans = 0;
      return ans;
    }
  }
  return ans;
}
int main() {
  int q;
  cin >> q;
  long long n;
  long long m;
  long long sum;
  while (q--) {
    cin >> n >> m;
    sum = 0;
    sum = ansew(n, m);
    cout << sum << endl;
  }
  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;
long long ansew(long long n, long long m) {
  long long ex;
  long long rep;
  long long ans = 0;
  ex = n / m;
  int undig = m % 10;
  switch (undig) {
    case 1: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += i;
      }
      return ans;
    }
    case 2: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (2 * i) % 10;
      }
      return ans;
    }
    case 3: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += ((3 * i) % 10);
      }
      return ans;
    }
    case 4: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (4 * i) % 10;
      }
      return ans;
    }
    case 5: {
      rep = ex / 2;
      ans += (5 * rep);
      for (int i = 1; i <= ex % 2; i++) {
        ans += (5 * i) % 10;
      }
      return ans;
    }
    case 6: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (6 * i) % 10;
      }
      return ans;
    }
    case 7: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += ((7 * i) % 10);
      }
      return ans;
    }
    case 8: {
      rep = ex / 5;
      ans += (20 * rep);
      for (int i = 1; i <= ex % 5; i++) {
        ans += (8 * i) % 10;
      }
      return ans;
    }
    case 9: {
      rep = ex / 10;
      ans += (45 * rep);
      for (int i = 1; i <= ex % 10; i++) {
        ans += (9 * i) % 10;
      }
      return ans;
    }
    case 0: {
      ans = 0;
      return ans;
    }
  }
  return ans;
}
int main() {
  int q;
  cin >> q;
  long long n;
  long long m;
  long long sum;
  while (q--) {
    cin >> n >> m;
    sum = 0;
    sum = ansew(n, m);
    cout << sum << endl;
  }
  return 0;
}

```