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;
int inf = INT_MAX;
long long modulo(long long num, long long MOD = 1000000007) {
  return ((num % MOD) + MOD) % MOD;
}
long long power(long long b, long long e, long long MOD = 1000000007) {
  long long ans = 1;
  while (e) {
    if (e % 2) ans = (ans * b) % MOD;
    b = (b * b) % MOD;
    e /= 2;
  }
  return ans;
}
long long inv(long long num, long long MOD = 1000000007) {
  return power(modulo(num), MOD - 2, MOD);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int q;
  long long n, m;
  cin >> q;
  while (q--) {
    cin >> n >> m;
    vector<int> dig;
    dig.push_back(m % 10);
    int first = m % 10;
    int curr = first;
    while (1) {
      curr = (curr + m) % 10;
      if (curr == first) break;
      dig.push_back(curr);
    }
    long long no = n / m;
    long long summ = 0;
    for (int num : dig) summ += num;
    long long ans = summ * (no / dig.size());
    int rem = no % dig.size();
    for (int i = 0; i < rem; i++) ans += dig[i];
    cout << ans << endl;
  }
  return 0;
}
