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

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

int N;
int A[200010];
int mem[30][50];

signed main()
{
    cin >> N;
    rep(i, N)
    {
        double d;
        cin >> d;
        A[i] = llround(d * 1e9);
    }

    int ans = 0;
    rep(i, N)
    {
        int cnt2 = 0, cnt5 = 0;
        int a = A[i];
        while(a % 2 == 0)
        {
            a /= 2;
            cnt2++;
        }
        while(a % 5 == 0)
        {
            a /= 5;
            cnt5++;
        }

        for(int j = 18 - cnt5; j < 30; j++)for(int k = 18 - cnt2; k < 50; k++)ans += mem[j][k];
        mem[cnt5][cnt2]++;
    }

    cout << ans <<endl;

    return 0;
}

### Prompt

Please create a solution in CPP 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 int long long
#define rep(i,n) for(int i=0;i<(n);i++)

int N;
int A[200010];
int mem[30][50];

signed main()
{
    cin >> N;
    rep(i, N)
    {
        double d;
        cin >> d;
        A[i] = llround(d * 1e9);
    }

    int ans = 0;
    rep(i, N)
    {
        int cnt2 = 0, cnt5 = 0;
        int a = A[i];
        while(a % 2 == 0)
        {
            a /= 2;
            cnt2++;
        }
        while(a % 5 == 0)
        {
            a /= 5;
            cnt5++;
        }

        for(int j = 18 - cnt5; j < 30; j++)for(int k = 18 - cnt2; k < 50; k++)ans += mem[j][k];
        mem[cnt5][cnt2]++;
    }

    cout << ans <<endl;

    return 0;
}

```