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

long a[222222], A[222222], d[111][111];
int main() {
    int n;
    cin >> n;
    for(int i = 0; i < n; i++) {
        double x;
        cin >> x;
        long b = round(x * 1e9);
        int c = 0;
        while(b % 2 == 0)
            c++, b /= 2;
        a[i] = c;
        int cc = 0;
        while(b % 5 == 0)
            cc++, b /= 5;
        A[i] = cc;
        d[c][cc]++;
    }
    for(int i = 100; i--;)
        for(int j = 100; j--;)
            d[i][j] += d[i + 1][j];
    for(int i = 100; i--;)
        for(int j = 100; j--;)
            d[i][j] += d[i][j + 1];
    long r = 0;
    for(int i = 0; i < n; i++) {
        int v = a[i];
        int u = max(18 - v, 0);
        int vv = A[i];
        int uu = max(18 - vv, 0);
        int x = d[u][uu];
        if(v >= u && vv >= uu)
            x--;
        r += x;
    }
    cout << r / 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;

long a[222222], A[222222], d[111][111];
int main() {
    int n;
    cin >> n;
    for(int i = 0; i < n; i++) {
        double x;
        cin >> x;
        long b = round(x * 1e9);
        int c = 0;
        while(b % 2 == 0)
            c++, b /= 2;
        a[i] = c;
        int cc = 0;
        while(b % 5 == 0)
            cc++, b /= 5;
        A[i] = cc;
        d[c][cc]++;
    }
    for(int i = 100; i--;)
        for(int j = 100; j--;)
            d[i][j] += d[i + 1][j];
    for(int i = 100; i--;)
        for(int j = 100; j--;)
            d[i][j] += d[i][j + 1];
    long r = 0;
    for(int i = 0; i < n; i++) {
        int v = a[i];
        int u = max(18 - v, 0);
        int vv = A[i];
        int uu = max(18 - vv, 0);
        int x = d[u][uu];
        if(v >= u && vv >= uu)
            x--;
        r += x;
    }
    cout << r / 2 << endl;
    return 0;
}

```