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

const int maxn = 200010;
int n, s1[maxn], s2[maxn], cnt[100][100];
long long a[maxn];

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        long double x; cin >> x;
        a[i] = round(x * 1e9);
        while (!(a[i] % 2)) s1[i]++, a[i] /= 2;
        while (!(a[i] % 5)) s2[i]++, a[i] /= 5;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 60; j++) {
            for (int k = 0; k <= 60; k++) {
                if (j + s1[i] >= 18 && k + s2[i] >= 18) ans += cnt[j][k];
            }
        }
        cnt[s1[i]][s2[i]]++;
    }
    cout << ans << endl;
    return 0;
}
### Prompt

Please formulate 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;

const int maxn = 200010;
int n, s1[maxn], s2[maxn], cnt[100][100];
long long a[maxn];

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        long double x; cin >> x;
        a[i] = round(x * 1e9);
        while (!(a[i] % 2)) s1[i]++, a[i] /= 2;
        while (!(a[i] % 5)) s2[i]++, a[i] /= 5;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 60; j++) {
            for (int k = 0; k <= 60; k++) {
                if (j + s1[i] >= 18 && k + s2[i] >= 18) ans += cnt[j][k];
            }
        }
        cnt[s1[i]][s2[i]]++;
    }
    cout << ans << endl;
    return 0;
}
```