#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

int ten[20][20];

int main() {
    cin.tie(0)->sync_with_stdio(false);

	int n; cin >> n;

	ll cnt = 0;
	REP(i, n) {
		double x; cin >> x;
		ll tmp = x * 1e9 + 0.1;
		int two = 0;
		int five = 0;
		while (tmp % 2 == 0) {
			tmp /= 2;
			two++;
		}
		while (tmp % 5 == 0) {
			tmp /= 5;
			five++;
		}
		if (two > 18) two = 18;
		if (five > 18) five = 18;
		for (int j = 18 - two; j <= 18; j++)
			for (int k = 18 - five; k <= 18; k++)
				cnt += ten[j][k];
		ten[two][five]++;
	}

	cout << cnt << '\n';
    return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

int ten[20][20];

int main() {
    cin.tie(0)->sync_with_stdio(false);

	int n; cin >> n;

	ll cnt = 0;
	REP(i, n) {
		double x; cin >> x;
		ll tmp = x * 1e9 + 0.1;
		int two = 0;
		int five = 0;
		while (tmp % 2 == 0) {
			tmp /= 2;
			two++;
		}
		while (tmp % 5 == 0) {
			tmp /= 5;
			five++;
		}
		if (two > 18) two = 18;
		if (five > 18) five = 18;
		for (int j = 18 - two; j <= 18; j++)
			for (int k = 18 - five; k <= 18; k++)
				cnt += ten[j][k];
		ten[two][five]++;
	}

	cout << cnt << '\n';
    return 0;
}

```