#include<bits/stdc++.h>
#define MAXN 100005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n;
vector<P> v;
int cnt[20][20];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        double x;
        scanf("%lf",&x);
        ll res=(ll)(x*INF+0.5);
        int cnt2=-9,cnt5=-9;
        while(res%2==0) res/=2,cnt2++;
        while(res%5==0) res/=5,cnt5++;
        cnt2=min(cnt2,9); cnt5=min(cnt5,9);
        v.push_back(P(cnt2,cnt5));
    }
    ll ans=0;
    for(int i=0;i<n;i++)
    {
        int x=v[i].F,y=v[i].S;
        for(int j=-9;j<=9;j++)
            for(int k=-9;k<=9;k++)
            {
                if(j+x<0||k+y<0) continue;
                ans+=cnt[j+9][k+9];
            }
        cnt[x+9][y+9]++;
    }
    printf("%lld\n",ans);
    return 0;
}
### Prompt

Your challenge is to write 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>
#define MAXN 100005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n;
vector<P> v;
int cnt[20][20];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        double x;
        scanf("%lf",&x);
        ll res=(ll)(x*INF+0.5);
        int cnt2=-9,cnt5=-9;
        while(res%2==0) res/=2,cnt2++;
        while(res%5==0) res/=5,cnt5++;
        cnt2=min(cnt2,9); cnt5=min(cnt5,9);
        v.push_back(P(cnt2,cnt5));
    }
    ll ans=0;
    for(int i=0;i<n;i++)
    {
        int x=v[i].F,y=v[i].S;
        for(int j=-9;j<=9;j++)
            for(int k=-9;k<=9;k++)
            {
                if(j+x<0||k+y<0) continue;
                ans+=cnt[j+9][k+9];
            }
        cnt[x+9][y+9]++;
    }
    printf("%lld\n",ans);
    return 0;
}
```