### Problem Description：

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 Specification:

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 Specification:

For each query print the answer for it — the sum of digits written down by Polycarp.

### Sample Input:

7
1 1
10 1
100 3
1024 14
998244353 1337
123 144
1234312817382646 13

1
45
153
294
3359835
427262129093995

### AC代码：TLE代码：

``````#include <bits/stdc++.h>
using namespace std;
#define Up(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;

int main()
{
ios::sync_with_stdio(false);
int q;
cin >> q;
while(q--)
{
ll n, m, sum = 0;
cin >> n >> m;
Up(i,1,n)
{
if(i%m == 0)
{
sum += i%10;
}
}
cout << sum << endl;
}
return 0;
}``````

### AC代码：

``````#include <bits/stdc++.h>
using namespace std;
#define Up(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
int main()
{
ios::sync_with_stdio(false);
int q;
cin >> q;
while(q--)
{
ll n, m, sum = 0;
cin >> n >> m;
ll num = n/m;
Up(i,1,9)
{
ll d = (m*i)%10;  //d表示m可能的个位数
ll cnt = num/10 + (num%10 >= i);  //cnt表示以d结尾的、能整除m的数的个数
sum += cnt*d;
}
cout << sum << endl;
}
return 0;
}``````