#include <bits/stdc++.h>
using namespace std;
#define reps(i, a, n) for (int i = (a); i < (n); ++i)
#define int long long

signed main(void)
{
	int n, ans = 0, num[20][20] = {0};
	cin >> n;
	while (n--)
	{
		double x; cin >> x;
		int tmp = x * 1e9 + 0.5;
		int n2 = 0, n5 = 0;
		while (tmp % 2 == 0) {n2++; tmp /= 2;}
		while (tmp % 5 == 0) {n5++; tmp /= 5;}
		if (n2 > 18) n2 = 18;
		if (n5 > 18) n5 = 18;
		reps (i, 18 - n2, 19)
		{
			reps (j, 18 - n5, 19)
			{
				ans += num[i][j];
			}
		}
		num[n2][n5]++;
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Please create a solution in CPP to 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 reps(i, a, n) for (int i = (a); i < (n); ++i)
#define int long long

signed main(void)
{
	int n, ans = 0, num[20][20] = {0};
	cin >> n;
	while (n--)
	{
		double x; cin >> x;
		int tmp = x * 1e9 + 0.5;
		int n2 = 0, n5 = 0;
		while (tmp % 2 == 0) {n2++; tmp /= 2;}
		while (tmp % 5 == 0) {n5++; tmp /= 5;}
		if (n2 > 18) n2 = 18;
		if (n5 > 18) n5 = 18;
		reps (i, 18 - n2, 19)
		{
			reps (j, 18 - n5, 19)
			{
				ans += num[i][j];
			}
		}
		num[n2][n5]++;
	}
	cout << ans << endl;
	return 0;
}

```