#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)

int main() {
	int N;
	long long p;
	long long aa;
 	double a0;
 	int a2;
 	int a5;
	vector<vector<int>> x(25, vector<int>(25));
	p=0;

	cin >> N;

   	rep(i,N) {
		cin >> a0;
		aa=a0*1e9+0.5;
   		rep(j,19) {
	        a2=j;
			if (aa%2>0) break;
			aa/=2;
		}
   		rep(j,19) {
	        a5=j;
			if (aa%5>0) break;
			aa/=5;
		}
	   	rep2(i1,18-a2,20) {
		   	rep2(j1,18-a5,20) p+=x.at(i1).at(j1);
		}
		x.at(a2).at(a5)++;
	}

	cout << p << endl;
	return 0;
}

### Prompt

Create a solution in CPP for the following problem:

You are given N real values A_1, A_2, \ldots, A_N. Compute the number of pairs of indices (i, j) such that i < j and the product A_i \cdot A_j is integer.

Constraints

* 2 \leq N \leq 200\,000
* 0 < A_i < 10^4
* A_i is given with at most 9 digits after the decimal.

Input

Input is given from Standard Input in the following format.


N
A_1
A_2
\vdots
A_N


Output

Print the number of pairs with integer product A_i \cdot A_j (and i < j).

Examples

Input

5
7.5
2.4
17.000000001
17
16.000000000


Output

3


Input

11
0.9
1
1
1.25
2.30000
5
70
0.000000001
9999.999999999
0.999999999
1.000000001


Output

8

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)

int main() {
	int N;
	long long p;
	long long aa;
 	double a0;
 	int a2;
 	int a5;
	vector<vector<int>> x(25, vector<int>(25));
	p=0;

	cin >> N;

   	rep(i,N) {
		cin >> a0;
		aa=a0*1e9+0.5;
   		rep(j,19) {
	        a2=j;
			if (aa%2>0) break;
			aa/=2;
		}
   		rep(j,19) {
	        a5=j;
			if (aa%5>0) break;
			aa/=5;
		}
	   	rep2(i1,18-a2,20) {
		   	rep2(j1,18-a5,20) p+=x.at(i1).at(j1);
		}
		x.at(a2).at(a5)++;
	}

	cout << p << endl;
	return 0;
}

```