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>
using namespace std;
bool is_powertwo(long long n) {
  long long count = 0;
  while (n) {
    if ((n & 1) > 0) count++;
    n = n >> 1;
  }
  if (count == 1) return true;
  return false;
}
void clear(std::queue<char> &q) {
  std::queue<char> empty;
  std::swap(q, empty);
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool is_prime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  long long t1 = t;
  while (t--) {
    long long n, m;
    cin >> n;
    cin >> m;
    long long arr[11];
    arr[0] = 0;
    for (long long i = 1; i <= 10; i++) {
      arr[i] = (i * m) % 10;
    }
    for (long long i = 1; i <= 10; i++) {
      arr[i] += arr[i - 1];
    }
    long long k = n / m;
    long long sum = (k / 10) * arr[10];
    long long k1 = k - (k / 10) * 10;
    sum += arr[k1];
    cout << sum << endl;
  }
  return 0;
}
