#include<bits/stdc++.h>
#define F first
#define S second
#define pb push_back
using namespace std;
const int maxn=1e6+3;
const int M=998244353;
typedef long long ll;
typedef pair<int,int> pi;
void add(int &x,int y){x+=y;if(x>=M)x-=M;}
void sub(int &x,int y){x-=y;if(x<0) x+=M;}
int n,c[100][100];
ll ans;
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++){
        double x; scanf("%lf",&x);
        ll y=(long long)(x*1000000000+0.5);
        int p2=0,p5=0;
        while (y%2==0) p2++,y/=2;
        while (y%5==0) p5++,y/=5;
        p2=min(p2,18); p5=min(p5,18);
        for (int j=18-p2;j<=18;j++)
            for (int k=18-p5;k<=18;k++)
                ans+=c[j][k];
        c[p2][p5]++;
    }
    printf("%lld\n",ans);
    return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 F first
#define S second
#define pb push_back
using namespace std;
const int maxn=1e6+3;
const int M=998244353;
typedef long long ll;
typedef pair<int,int> pi;
void add(int &x,int y){x+=y;if(x>=M)x-=M;}
void sub(int &x,int y){x-=y;if(x<0) x+=M;}
int n,c[100][100];
ll ans;
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++){
        double x; scanf("%lf",&x);
        ll y=(long long)(x*1000000000+0.5);
        int p2=0,p5=0;
        while (y%2==0) p2++,y/=2;
        while (y%5==0) p5++,y/=5;
        p2=min(p2,18); p5=min(p5,18);
        for (int j=18-p2;j<=18;j++)
            for (int k=18-p5;k<=18;k++)
                ans+=c[j][k];
        c[p2][p5]++;
    }
    printf("%lld\n",ans);
    return 0;
}

```