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

#define ll long long
#define rep(i,n) for(int i = 0;i < (n); i++)

int main() {
	ll n;
	ll i, j;
	cin >> n;
	ll a[19][19] = { 0 };
	ll sum[20][20] = { 0 };
	double d;
	ll buf;
	ll ans = 0;
	int buf2, buf5;

	rep(i, n) {
		cin >> d;
		buf = llround(d * 1e9);
		buf2 = 0;		buf5 = 0;
		while (buf % 2 == 0) {
			buf2++;
			buf /= 2;
		}
		while (buf % 5 == 0) {
			buf5++;
			buf /= 5;
		}
		if (buf2 > 18)buf2 = 18;
		if (buf5 > 18)buf5 = 18;
		a[buf2][buf5]++;
	}
	for (i = 18; i >= 0; i--) {
		for (j = 18; j >= 0; j--) {
			sum[i][j] = a[i][j] + sum[i + 1][j] + sum[i][j + 1] - sum[i + 1][j + 1];
		}
	}
	for (i = 18; i >= 0; i-- ) {
		for (j = 18; j >= 0; j--) {
			if (i >= 18 - i && j >= 18 - j) {
				ans += a[i][j] * (sum[18 - i][18 - j] - 1);
			}
			else {
				ans += a[i][j] * (sum[18 - i][18 - j]);
			}
			
		}
	}
	cout << ans / 2 << endl;

	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;

#define ll long long
#define rep(i,n) for(int i = 0;i < (n); i++)

int main() {
	ll n;
	ll i, j;
	cin >> n;
	ll a[19][19] = { 0 };
	ll sum[20][20] = { 0 };
	double d;
	ll buf;
	ll ans = 0;
	int buf2, buf5;

	rep(i, n) {
		cin >> d;
		buf = llround(d * 1e9);
		buf2 = 0;		buf5 = 0;
		while (buf % 2 == 0) {
			buf2++;
			buf /= 2;
		}
		while (buf % 5 == 0) {
			buf5++;
			buf /= 5;
		}
		if (buf2 > 18)buf2 = 18;
		if (buf5 > 18)buf5 = 18;
		a[buf2][buf5]++;
	}
	for (i = 18; i >= 0; i--) {
		for (j = 18; j >= 0; j--) {
			sum[i][j] = a[i][j] + sum[i + 1][j] + sum[i][j + 1] - sum[i + 1][j + 1];
		}
	}
	for (i = 18; i >= 0; i-- ) {
		for (j = 18; j >= 0; j--) {
			if (i >= 18 - i && j >= 18 - j) {
				ans += a[i][j] * (sum[18 - i][18 - j] - 1);
			}
			else {
				ans += a[i][j] * (sum[18 - i][18 - j]);
			}
			
		}
	}
	cout << ans / 2 << endl;

	return 0;
}
```