#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    if (n < m) {
      cout << 0 << endl;
      continue;
    }
    map<long long, long long> mp;
    long long s = 0;
    vector<long long> ar;
    for (long long i = m; i <= n; i += m) {
      long long x = i % 10;
      if (mp[x]) break;
      mp[x] = 1;
      ar.push_back(x);
      s += x;
    }
    long long r = n / m;
    long long ans = (r / ar.size()) * s;
    for (int i = 0; i < r % ar.size(); i++) ans += ar[i];
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    if (n < m) {
      cout << 0 << endl;
      continue;
    }
    map<long long, long long> mp;
    long long s = 0;
    vector<long long> ar;
    for (long long i = m; i <= n; i += m) {
      long long x = i % 10;
      if (mp[x]) break;
      mp[x] = 1;
      ar.push_back(x);
      s += x;
    }
    long long r = n / m;
    long long ans = (r / ar.size()) * s;
    for (int i = 0; i < r % ar.size(); i++) ans += ar[i];
    cout << ans << endl;
  }
  return 0;
}

```