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

const int D = 1e9;
const int SIZE = 2e5 + 5;
int n;
struct P{
ll dat;
int c[4];
} a[SIZE], d[SIZE];
int cnt[50][50];

int main()
{
	double b;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
	{
		scanf("%lf", &b);
		a[i].dat = llround(b * D);
		ll p = a[i].dat;
		while (p % 2 == 0)
		{
			++a[i].c[2];  p /= 2;
		}
		while (p % 5 == 0)
		{
			++a[i].c[3]; p /= 5;
		}
		++cnt[a[i].c[2]][a[i].c[3]];
	}
	ll ans = 0;
	for (int i = 0; i <= 49; ++i)
	{
		for (int j = 0; j <= 49; ++j)
		{
			for (int k = 0; k <= 49; ++k)
			{
				for (int l = 0; l <= 49; ++l)
				{
					if (min(i + k, j + l) >= 18)
					{
						if (i == k && j == l) ans += (ll)cnt[i][j] * (cnt[k][l] - 1);
						else ans += (ll)cnt[i][j] * cnt[k][l];
					}
				}
			}
		}
	}
	printf("%lld\n", ans >> 1);
	return 0;
}
### Prompt

Your challenge is to write 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;
typedef long long ll;

const int D = 1e9;
const int SIZE = 2e5 + 5;
int n;
struct P{
ll dat;
int c[4];
} a[SIZE], d[SIZE];
int cnt[50][50];

int main()
{
	double b;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
	{
		scanf("%lf", &b);
		a[i].dat = llround(b * D);
		ll p = a[i].dat;
		while (p % 2 == 0)
		{
			++a[i].c[2];  p /= 2;
		}
		while (p % 5 == 0)
		{
			++a[i].c[3]; p /= 5;
		}
		++cnt[a[i].c[2]][a[i].c[3]];
	}
	ll ans = 0;
	for (int i = 0; i <= 49; ++i)
	{
		for (int j = 0; j <= 49; ++j)
		{
			for (int k = 0; k <= 49; ++k)
			{
				for (int l = 0; l <= 49; ++l)
				{
					if (min(i + k, j + l) >= 18)
					{
						if (i == k && j == l) ans += (ll)cnt[i][j] * (cnt[k][l] - 1);
						else ans += (ll)cnt[i][j] * cnt[k][l];
					}
				}
			}
		}
	}
	printf("%lld\n", ans >> 1);
	return 0;
}
```