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
#include <bits/stdc++.h>
const long long int M = 1000000000;
using namespace std;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, m;
    cin >> n >> m;
    long long int x = m % 10;
    if (m > n) {
      cout << "0\n";
      continue;
    }
    if (x == 0) {
      cout << "0\n";
      continue;
    } else if (x == 5) {
      long long int sum = 0;
      for (long long int i = 1; i <= min(2LL, n / m); i++) {
        int c = ((i * x) % 10);
        sum += c;
      }
      if (n / m <= 2)
        cout << sum << "\n";
      else {
        long long int z = (n / m);
        long long int k = z / 2;
        long long int r = z % 2;
        sum = sum * k;
        for (long long int i = 1; i <= r; i++) {
          int c = ((i * x) % 10);
          sum += c;
        }
        cout << sum << "\n";
      }
    } else if (x % 2 == 0) {
      long long int sum = 0;
      for (long long int i = 1; i <= min(5LL, n / m); i++) {
        int c = ((i * x) % 10);
        sum += c;
      }
      if (n / m < 5)
        cout << sum << "\n";
      else {
        long long int z = (n / m);
        long long int k = z / 5;
        long long int r = z % 5;
        sum = sum * k;
        for (long long int i = 1; i <= r; i++) {
          int c = ((i * x) % 10);
          sum += c;
        }
        cout << sum << "\n";
      }
    } else {
      long long int sum = 0;
      for (long long int i = 1; i <= min(10LL, n / m); i++) {
        int c = ((i * x) % 10);
        sum += c;
      }
      if (n / m < 10)
        cout << sum << "\n";
      else {
        long long int z = (n / m);
        long long int k = z / 10;
        long long int r = z % 10;
        sum = sum * k;
        for (long long int i = 1; i <= r; i++) {
          int c = ((i * x) % 10);
          sum += c;
        }
        cout << sum << "\n";
      }
    }
  }
  return 0;
}
