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

int64_t cnt[50][50];
int main() {
	int n; cin >> n;
	int64_t a[n];
	for(int i = 0; i < n; ++i) {
		double aa; cin >> aa;
		a[i] = llround(aa * 1e9);
		int t = 0;
		while(a[i] % 2 == 0) {
			a[i] /= 2;
			t++;
		}
		int f = 0;
		while(a[i] % 5 == 0) {
			a[i] /= 5;
			f++;
		}
		++cnt[t][f];
	}
	int64_t ans = 0;
	for(int i = 0; i < 50 * 50; ++i) {
		int il = i / 50, ir = i % 50;
		for(int j = i; j < 50 * 50; ++j) {
			int jl = j / 50, jr = j % 50;
			if(il + jl >= 18 and ir + jr >= 18) {
				if(i == j) {
					ans += cnt[il][ir] * (cnt[il][ir] - 1) / 2;
				} else {
					ans += cnt[il][ir] * cnt[jl][jr];
				}
			}
		}
	}
	cout << ans << '\n';
	return 0;
}
### Prompt

Generate a cpp solution 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>
using namespace std;

int64_t cnt[50][50];
int main() {
	int n; cin >> n;
	int64_t a[n];
	for(int i = 0; i < n; ++i) {
		double aa; cin >> aa;
		a[i] = llround(aa * 1e9);
		int t = 0;
		while(a[i] % 2 == 0) {
			a[i] /= 2;
			t++;
		}
		int f = 0;
		while(a[i] % 5 == 0) {
			a[i] /= 5;
			f++;
		}
		++cnt[t][f];
	}
	int64_t ans = 0;
	for(int i = 0; i < 50 * 50; ++i) {
		int il = i / 50, ir = i % 50;
		for(int j = i; j < 50 * 50; ++j) {
			int jl = j / 50, jr = j % 50;
			if(il + jl >= 18 and ir + jr >= 18) {
				if(i == j) {
					ans += cnt[il][ir] * (cnt[il][ir] - 1) / 2;
				} else {
					ans += cnt[il][ir] * cnt[jl][jr];
				}
			}
		}
	}
	cout << ans << '\n';
	return 0;
}
```