#include <bits/stdc++.h>

typedef long long ll;

const int E = 50;

int n, cnt[E][E];
ll ans;
char s[E];

int main()
{
	std::cin >> n;
	for (int i = 1; i <= n; i++)
	{
		scanf("%s", s + 1); int l = strlen(s + 1);
		ll tmp = 0; int c2 = -9, c5 = -9, c = 9; bool is = 0;
		for (int i = 1; i <= l; i++)
			if (s[i] == '.') {is = 1; continue;}
			else if (is) tmp = tmp * 10 + s[i] - '0', c--;
			else tmp = tmp * 10 + s[i] - '0';
		while (c--) tmp *= 10;
		while (tmp % 2 == 0) tmp /= 2, c2++;
		while (tmp % 5 == 0) tmp /= 5, c5++;
		for (int i = 36; i >= -c2 && i >= -9; i--)
			for (int j = 20; j >= -c5 && j >= -9; j--)
				ans += cnt[i + 9][j + 9];
		cnt[c2 + 9][c5 + 9]++;
	}
	return std::cout << ans << std::endl, 0;
}
### Prompt

Please create a solution in Cpp 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>

typedef long long ll;

const int E = 50;

int n, cnt[E][E];
ll ans;
char s[E];

int main()
{
	std::cin >> n;
	for (int i = 1; i <= n; i++)
	{
		scanf("%s", s + 1); int l = strlen(s + 1);
		ll tmp = 0; int c2 = -9, c5 = -9, c = 9; bool is = 0;
		for (int i = 1; i <= l; i++)
			if (s[i] == '.') {is = 1; continue;}
			else if (is) tmp = tmp * 10 + s[i] - '0', c--;
			else tmp = tmp * 10 + s[i] - '0';
		while (c--) tmp *= 10;
		while (tmp % 2 == 0) tmp /= 2, c2++;
		while (tmp % 5 == 0) tmp /= 5, c5++;
		for (int i = 36; i >= -c2 && i >= -9; i--)
			for (int j = 20; j >= -c5 && j >= -9; j--)
				ans += cnt[i + 9][j + 9];
		cnt[c2 + 9][c5 + 9]++;
	}
	return std::cout << ans << std::endl, 0;
}
```