#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
int main() {
	int N;
	scanf("%d", &N);
	double A[N];
	for (int i = 0; i < N; ++i) scanf("%lf", &A[i]);
	long long B[N], ans = 0;
	for (int i = 0; i < N; ++i) B[i] = llround(A[i] * 1000000000LL);
	vector<long long> p5[19];
	int p2[44];
	fill_n(p2, 44, 0);
	for (int i = 0; i < N; ++i) {
		int x = 0;
		long long temp = B[i];
		while (temp % 5 == 0) ++x, temp /= 5;
		p5[x].push_back(B[i]);
	}
	for (int i = 0; i < 19; ++i) {
		for (long long j: p5[18 - i]) {
			int x = 0;
			long long temp = j;
			while (temp % 2 == 0) ++x, temp /= 2;
			++p2[x];
		}
		for (long long j: p5[i]) {
			int x = 0;
			long long temp = j;
			while (temp % 2 == 0) ++x, temp /= 2;
			for (int k = max(0, 18 - x); k < 44; ++k) ans += p2[k];
			if (i * 2 >= 18 && x * 2 >= 18) --ans;
		}
	}
	printf("%lld", ans / 2);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
int main() {
	int N;
	scanf("%d", &N);
	double A[N];
	for (int i = 0; i < N; ++i) scanf("%lf", &A[i]);
	long long B[N], ans = 0;
	for (int i = 0; i < N; ++i) B[i] = llround(A[i] * 1000000000LL);
	vector<long long> p5[19];
	int p2[44];
	fill_n(p2, 44, 0);
	for (int i = 0; i < N; ++i) {
		int x = 0;
		long long temp = B[i];
		while (temp % 5 == 0) ++x, temp /= 5;
		p5[x].push_back(B[i]);
	}
	for (int i = 0; i < 19; ++i) {
		for (long long j: p5[18 - i]) {
			int x = 0;
			long long temp = j;
			while (temp % 2 == 0) ++x, temp /= 2;
			++p2[x];
		}
		for (long long j: p5[i]) {
			int x = 0;
			long long temp = j;
			while (temp % 2 == 0) ++x, temp /= 2;
			for (int k = max(0, 18 - x); k < 44; ++k) ans += p2[k];
			if (i * 2 >= 18 && x * 2 >= 18) --ans;
		}
	}
	printf("%lld", ans / 2);
	return 0;
}
```