#include <bits/stdc++.h>
using namespace std;

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int n;
	cin >> n;
	map<pair<int,int>, int> cnt;
	for(int rep = 0; rep < n; ++rep) {
		double x;
		cin >> x;
		long long i = llround(x * 1e9);
		int two = 0, five = 0;
		while(i % 2 == 0) {
			two++;
			i /= 2;
		}
		while(i % 5 == 0) {
			five++;
			i /= 5;
		}
		++cnt[{two,five}];
	}
	long long ans = 0;
	for (auto p1 : cnt) {
		for (auto p2 : cnt) {
			if (p1.first.first+p2.first.first >= 18 && p1.first.second+p2.first.second >= 18) {
				if (p1 < p2) {
					ans += (long long) p1.second * p2.second;
				} else if (p1 == p2) {
					ans += (long long) p1.second * (p1.second - 1) / 2;
				}
			}
		}
	}
	cout << ans << '\n';
	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
#include <bits/stdc++.h>
using namespace std;

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int n;
	cin >> n;
	map<pair<int,int>, int> cnt;
	for(int rep = 0; rep < n; ++rep) {
		double x;
		cin >> x;
		long long i = llround(x * 1e9);
		int two = 0, five = 0;
		while(i % 2 == 0) {
			two++;
			i /= 2;
		}
		while(i % 5 == 0) {
			five++;
			i /= 5;
		}
		++cnt[{two,five}];
	}
	long long ans = 0;
	for (auto p1 : cnt) {
		for (auto p2 : cnt) {
			if (p1.first.first+p2.first.first >= 18 && p1.first.second+p2.first.second >= 18) {
				if (p1 < p2) {
					ans += (long long) p1.second * p2.second;
				} else if (p1 == p2) {
					ans += (long long) p1.second * (p1.second - 1) / 2;
				}
			}
		}
	}
	cout << ans << '\n';
	return 0;
}
```