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;
clock_t present = clock();
void complexiti() {
  present = clock() - present;
  cerr << "TIME : " << float(present) / CLOCKS_PER_SEC << '\n';
}
void solver() {
  long long n, m;
  cin >> n >> m;
  vector<long long> v[10];
  for (int i = 0; i < 10; ++i) {
    int sum = i;
    while (sum % 10 != 0) {
      v[i].push_back(sum % 10);
      sum += i;
    }
    v[i].push_back(0);
  }
  long long times = n / m;
  long long rounds = times / v[m % 10].size();
  long long x = times % v[m % 10].size();
  long long res = 0, sum = 0;
  for (int i = 0; i < v[m % 10].size(); ++i) {
    sum += v[m % 10][i];
  }
  for (long long i = 0; i < x; ++i) {
    res += v[m % 10][i];
  }
  res += rounds * sum;
  cout << res << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int queries = 1;
  (cin >> queries);
  for (int i = 0; i < queries; ++i) {
    solver();
  }
  complexiti();
  return 0;
}
