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

int64_t cnt[41][41];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int n; cin >> n;
	for (int i = 0; i < n; i++) {
		double x; cin >> x;
		int64_t a = x * 1e9 + 0.5;
		int c1 = 0, c2 = 0;
		while (a % 5 == 0) { c1++; a /= 5; }
		while (a % 2 == 0) { c2++; a /= 2; }
		cnt[c1][c2]++;
	}
	int64_t ans = 0;
	for (int i = 0; i < 41; i++) {
		for (int j = 0; j < 41; j++) {
			for (int x = max(18 - i, 0); x < 41; x++) {
				for (int y = max(18 - j, 0); y < 41; y++) {
					if (x == i && y == j) {
						ans += cnt[i][j] * (cnt[x][y] - 1);
					} else {
						ans += cnt[i][j] * cnt[x][y];
					}
				}
			}
		}
	}
	cout << ans / 2;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;

int64_t cnt[41][41];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int n; cin >> n;
	for (int i = 0; i < n; i++) {
		double x; cin >> x;
		int64_t a = x * 1e9 + 0.5;
		int c1 = 0, c2 = 0;
		while (a % 5 == 0) { c1++; a /= 5; }
		while (a % 2 == 0) { c2++; a /= 2; }
		cnt[c1][c2]++;
	}
	int64_t ans = 0;
	for (int i = 0; i < 41; i++) {
		for (int j = 0; j < 41; j++) {
			for (int x = max(18 - i, 0); x < 41; x++) {
				for (int y = max(18 - j, 0); y < 41; y++) {
					if (x == i && y == j) {
						ans += cnt[i][j] * (cnt[x][y] - 1);
					} else {
						ans += cnt[i][j] * cnt[x][y];
					}
				}
			}
		}
	}
	cout << ans / 2;
}
```