#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  const int N = 50;
  vector<vector<int>> cnt(N,vector<int>(N));
  for (int i = 1; i <= n; i++) {
  	long double x ;
  	cin >> x;
  	long long a = (long long)(x * 1e9 + 0.5);
  	int n1 = 0,n2 = 0;
  	while (a % 2 == 0) {
  		a /= 2;
  		n1 += 1;
		}
		while (a % 5 == 0) {
			a /= 5;
			n2 += 1;
		}
		++cnt[n1][n2];
	}
	long long ans = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			for (int ii = 0; ii < N; ii++) {
				for (int jj = 0; jj < N; jj++) {
					if (min(i + ii,j + jj) >= 18) {
						if (i != ii || j != jj) {
							ans += 1ll * cnt[i][j] * cnt[ii][jj];
						}
						else {
							ans += 1ll * cnt[i][j] * (cnt[i][j] - 1);
						}
					}
				}
			}
		}
	}
	cout << ans / 2;
} 
### 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>
 
using namespace std;
 
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  const int N = 50;
  vector<vector<int>> cnt(N,vector<int>(N));
  for (int i = 1; i <= n; i++) {
  	long double x ;
  	cin >> x;
  	long long a = (long long)(x * 1e9 + 0.5);
  	int n1 = 0,n2 = 0;
  	while (a % 2 == 0) {
  		a /= 2;
  		n1 += 1;
		}
		while (a % 5 == 0) {
			a /= 5;
			n2 += 1;
		}
		++cnt[n1][n2];
	}
	long long ans = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			for (int ii = 0; ii < N; ii++) {
				for (int jj = 0; jj < N; jj++) {
					if (min(i + ii,j + jj) >= 18) {
						if (i != ii || j != jj) {
							ans += 1ll * cnt[i][j] * cnt[ii][jj];
						}
						else {
							ans += 1ll * cnt[i][j] * (cnt[i][j] - 1);
						}
					}
				}
			}
		}
	}
	cout << ans / 2;
} 
```