#include <bits/stdc++.h>
using namespace std;
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	map<pair<short,short>,int> cnt;
	for (int i = 0; i < n; i++) {
		double x;
		cin >> x;
		long long y = llround(x * 1e9);
		short two = 0, five = 0;
		while (!(y % 2)) {
			two++;
			y /= 2;
		}
		while (!(y % 5)) {
			five++;
			y /= 5;
		}
		cnt[pair<short, short>(two, five)]++;
	}
	long long ans = 0;
	for (map<pair<short, short>, int>::iterator i = cnt.begin(); i != cnt.end(); i++)
		for (map<pair<short, short>, int>::iterator j = cnt.begin(); j != cnt.end(); j++)
			if (i->first.first + j->first.first >= 18 && i->first.second + j->first.second >= 18)
				if (i->first < j->first) ans += 1ll * i->second * j->second;
				else if (i->first == j->first) ans += 1ll * i->second * (i->second - 1) / 2;
	cout << ans << "\n";
	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;
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	map<pair<short,short>,int> cnt;
	for (int i = 0; i < n; i++) {
		double x;
		cin >> x;
		long long y = llround(x * 1e9);
		short two = 0, five = 0;
		while (!(y % 2)) {
			two++;
			y /= 2;
		}
		while (!(y % 5)) {
			five++;
			y /= 5;
		}
		cnt[pair<short, short>(two, five)]++;
	}
	long long ans = 0;
	for (map<pair<short, short>, int>::iterator i = cnt.begin(); i != cnt.end(); i++)
		for (map<pair<short, short>, int>::iterator j = cnt.begin(); j != cnt.end(); j++)
			if (i->first.first + j->first.first >= 18 && i->first.second + j->first.second >= 18)
				if (i->first < j->first) ans += 1ll * i->second * j->second;
				else if (i->first == j->first) ans += 1ll * i->second * (i->second - 1) / 2;
	cout << ans << "\n";
	return 0;
}
```