#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef _MSC_VER
	freopen("a.in", "r", stdin);
#endif
	std::ios::sync_with_stdio(false);
	int n;
	cin >> n;
	int c[32][32];
	memset(c, 0, sizeof(c));
	long long ret = 0;
	const int L = 18;
	for (int i = 0; i < n; i++) {
		double val;
		cin >> val;
		long long d = (long long)(val * 1e9 + 0.5);
		int c2 = 0;
		int c5 = 0;
		for (; d % 2 == 0; d /= 2) ++c2;
		for (; d % 5 == 0; d /= 5) ++c5;
		c2 = min(c2, L);
		c5 = min(c5, L);
		for (int x2 = 0; x2 <= L; x2++) {
			if (c2 + x2 >= L) {
				ret += c[x2][L - c5];
			}
		}
		for (int x5 = 0; x5 <= c5; ++x5) {
			++c[c2][x5];
		}
	}
	cout << ret << endl;
	return 0;
}

### Prompt

Your task is to create a cpp solution 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
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef _MSC_VER
	freopen("a.in", "r", stdin);
#endif
	std::ios::sync_with_stdio(false);
	int n;
	cin >> n;
	int c[32][32];
	memset(c, 0, sizeof(c));
	long long ret = 0;
	const int L = 18;
	for (int i = 0; i < n; i++) {
		double val;
		cin >> val;
		long long d = (long long)(val * 1e9 + 0.5);
		int c2 = 0;
		int c5 = 0;
		for (; d % 2 == 0; d /= 2) ++c2;
		for (; d % 5 == 0; d /= 5) ++c5;
		c2 = min(c2, L);
		c5 = min(c5, L);
		for (int x2 = 0; x2 <= L; x2++) {
			if (c2 + x2 >= L) {
				ret += c[x2][L - c5];
			}
		}
		for (int x5 = 0; x5 <= c5; ++x5) {
			++c[c2][x5];
		}
	}
	cout << ret << endl;
	return 0;
}

```