#include <cstdio>
#include <iostream>
using namespace std;

#define N 200100
#define ll long long

int cnt[60][30];

int main()
{
	int n;
	ll ans = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		double x;
		scanf("%lf", &x);
		ll v = x * 1e9 + .5;
		int num2 = 0, num5 = 0;
		while (v % 2 == 0) {
			num2++;
			v /= 2;
		}
		while (v % 5 == 0) {
			num5++;
			v /= 5;
		}
		num2 = min(num2, 18);
		num5 = min(num5, 18);
		for (int i = 18 - num2; i <= 18; i++)
			for (int j = 18 - num5; j <= 18; j++)
				ans += cnt[i][j];
		cnt[num2][num5]++;
	}
	printf("%lld\n", ans);
}
### Prompt

Generate 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 <cstdio>
#include <iostream>
using namespace std;

#define N 200100
#define ll long long

int cnt[60][30];

int main()
{
	int n;
	ll ans = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		double x;
		scanf("%lf", &x);
		ll v = x * 1e9 + .5;
		int num2 = 0, num5 = 0;
		while (v % 2 == 0) {
			num2++;
			v /= 2;
		}
		while (v % 5 == 0) {
			num5++;
			v /= 5;
		}
		num2 = min(num2, 18);
		num5 = min(num5, 18);
		for (int i = 18 - num2; i <= 18; i++)
			for (int j = 18 - num5; j <= 18; j++)
				ans += cnt[i][j];
		cnt[num2][num5]++;
	}
	printf("%lld\n", ans);
}
```