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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long int n, m, q, s1 = 0, ans = 0;
vector<long long int> v;
int32_t main() {
  ios_base::sync_with_stdio();
  cin.tie();
  cout.tie();
  ;
  cout << fixed;
  cin >> q;
  while (q--) {
    cin >> n;
    cin >> m;
    ans = 0;
    v.clear();
    v.push_back(m % 10);
    s1 = n / m;
    long long int x = 1, s2 = m % 10;
    while ((x * s2) % 10 != 0) {
      x++;
      v.push_back((x * s2) % 10);
    }
    if (s1 % v.size() == 0) {
      x = s1 / v.size();
      for (long long int i = 0; i < v.size(); i++) ans = ans + v[i] * x;
    } else {
      x = s1 / v.size();
      for (long long int i = 0; i < s1 % v.size(); i++) {
        ans = ans + v[i];
      }
      for (long long int i = 0; i < v.size(); i++) {
        ans = ans + x * v[i];
      }
    }
    cout << ans << endl;
  }
}
