/* be name Khoda */
#include <bits/stdc++.h>
using namespace std;

const int N = 200 * 1001, inf = 1e9;
int n;
long long a[N], res;
long double d[N];
map<pair<int, int>, int> cnt;

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> d[i];
		a[i] = llround(d[i] * inf); 
		int two = 0, five = 0;
		while (a[i] % 2 == 0) {
			two++;
			a[i] /= 2;
		}
		while (a[i] % 5 == 0) {
			five++;
			a[i] /= 5;
		}
		cnt[{two, five}]++;
	}
	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)
					res += 1LL * p1.second * p2.second;
				else if (p1 == p2)
					res += 1LL * p1.second * (p1.second - 1) / 2;
			}
		}
	cout << res << "\n";
}

### Prompt

In cpp, your task is to solve 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
/* be name Khoda */
#include <bits/stdc++.h>
using namespace std;

const int N = 200 * 1001, inf = 1e9;
int n;
long long a[N], res;
long double d[N];
map<pair<int, int>, int> cnt;

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> d[i];
		a[i] = llround(d[i] * inf); 
		int two = 0, five = 0;
		while (a[i] % 2 == 0) {
			two++;
			a[i] /= 2;
		}
		while (a[i] % 5 == 0) {
			five++;
			a[i] /= 5;
		}
		cnt[{two, five}]++;
	}
	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)
					res += 1LL * p1.second * p2.second;
				else if (p1 == p2)
					res += 1LL * p1.second * (p1.second - 1) / 2;
			}
		}
	cout << res << "\n";
}

```