#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int n;
    ll a[69][69] = {}, z = 0;
    double d;
    cin >> n;
    while (n--) {
        cin >> d;
        ll b = (ll)round(d * 1e9);
        int x = 0, y = 0;
        while (b % 2 == 0) b /= 2, x++;
        while (b % 5 == 0) b /= 5, y++;
        a[x][y]++;
        //cout << x << ' ' << y << '\n';
    }
    for (int i = 0; i < 69; i++) for (int j = 0; j < 69; j++) {
        for (int k = 0; k < 69; k++) for (int l = 0; l < 69; l++) {
            if (i + k >= 18 && j + l >= 18) {
                if (i == k && j == l) z += a[i][j] * (a[i][j] - 1) / 2;
                else if (i < k || (i == k && j < l)) z += a[i][j] * a[k][l];
            }
        }
    }
    cout << z << '\n';
}

### 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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int n;
    ll a[69][69] = {}, z = 0;
    double d;
    cin >> n;
    while (n--) {
        cin >> d;
        ll b = (ll)round(d * 1e9);
        int x = 0, y = 0;
        while (b % 2 == 0) b /= 2, x++;
        while (b % 5 == 0) b /= 5, y++;
        a[x][y]++;
        //cout << x << ' ' << y << '\n';
    }
    for (int i = 0; i < 69; i++) for (int j = 0; j < 69; j++) {
        for (int k = 0; k < 69; k++) for (int l = 0; l < 69; l++) {
            if (i + k >= 18 && j + l >= 18) {
                if (i == k && j == l) z += a[i][j] * (a[i][j] - 1) / 2;
                else if (i < k || (i == k && j < l)) z += a[i][j] * a[k][l];
            }
        }
    }
    cout << z << '\n';
}

```