#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < ll(n); i++)

const ll mod = 1e9;
ll counter[20][20];

int main() {
    int n;
    cin >> n;
    ll ans = 0;
    while (n--) {
        double a;
        cin >> a;
        ll val = a * mod + 0.5;
        int n2 = 1, n5 = 1;
        while (val > 0 && val % 2 == 0) {
            val /= 2;
            n2++;
        }
        while (val > 0 && val % 5 == 0) {
            val /= 5;
            n5++;
        }
        n2 = min(n2, 19);
        n5 = min(n5, 19);
        for (int i = 20 - n2; i <= 19; i++) {
            for (int j = 20 - n5; j <= 19; j++) {
                ans += counter[i][j];
            }
        }
        counter[n2][n5]++;
    }
    cout << ans << endl;
}
### 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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < ll(n); i++)

const ll mod = 1e9;
ll counter[20][20];

int main() {
    int n;
    cin >> n;
    ll ans = 0;
    while (n--) {
        double a;
        cin >> a;
        ll val = a * mod + 0.5;
        int n2 = 1, n5 = 1;
        while (val > 0 && val % 2 == 0) {
            val /= 2;
            n2++;
        }
        while (val > 0 && val % 5 == 0) {
            val /= 5;
            n5++;
        }
        n2 = min(n2, 19);
        n5 = min(n5, 19);
        for (int i = 20 - n2; i <= 19; i++) {
            for (int j = 20 - n5; j <= 19; j++) {
                ans += counter[i][j];
            }
        }
        counter[n2][n5]++;
    }
    cout << ans << endl;
}
```