#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int maxn = 2e5 + 5;
const int mod = 998244353ll;
const long long INF = 1e18;
const int MX = 2000000001;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long t, q, n, a, b, c, k = 0, l, m, r, x = 0, y = 0, sum = 0, ans = 0,
                              temp = 0, res = 0;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    vector<long long> v(11);
    for (long long i = 1; i <= 10; i++) v[i] = (m * i) % 10;
    temp = n / m;
    q = temp / 10;
    r = temp % 10;
    ans = 0;
    for (long long i = 1; i < (11); i++) ans += v[i];
    ans *= q;
    for (long long i = 1; i < (r + 1); i++) ans += v[i];
    cout << ans << "\n";
  }
}

### Prompt

Your task is to create a CPP solution to 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int maxn = 2e5 + 5;
const int mod = 998244353ll;
const long long INF = 1e18;
const int MX = 2000000001;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long t, q, n, a, b, c, k = 0, l, m, r, x = 0, y = 0, sum = 0, ans = 0,
                              temp = 0, res = 0;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    vector<long long> v(11);
    for (long long i = 1; i <= 10; i++) v[i] = (m * i) % 10;
    temp = n / m;
    q = temp / 10;
    r = temp % 10;
    ans = 0;
    for (long long i = 1; i < (11); i++) ans += v[i];
    ans *= q;
    for (long long i = 1; i < (r + 1); i++) ans += v[i];
    cout << ans << "\n";
  }
}

```