#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
ll a[200005],b[200005],cnt[200005],A[2005],B[2005],ans=0;
int main() {
	cin>>n;
	for(int i=1;i<=n;i++){
		long double x;
		scanf("%Lf",&x),a[i]=round(x*1e9);
		if(a[i]%1000000000==0)ans--;
	}
	for(ll i=0,p2=1;p2<=1000000000000000000ll;i++,p2<<=1){
		for(ll j=0,p5=1;p2*p5<=1000000000000000000ll;j++,p5*=5)b[++b[0]]=p2*p5,A[b[0]]=i,B[b[0]]=j;
	}
	for(ll i=1;i<=n;i++){
		for(ll j=1;j<=b[0];j++){
			if(a[i]%b[j]==0&&a[i]%(b[j]*2)!=0&&a[i]%(b[j]*5)!=0){
				cnt[j]++;//cout<<b[j]<<' '<<A[j]<<' '<<B[j]<<endl;
				break;
			}
		}
	}
	for(int i=1;i<=b[0];i++){
		for(int j=1;j<=b[0];j++)if(A[i]+A[j]>=18&&B[i]+B[j]>=18)ans+=cnt[i]*cnt[j];
	}
	cout<<ans/2<<endl;
	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>
using namespace std;
typedef long long ll;
int n;
ll a[200005],b[200005],cnt[200005],A[2005],B[2005],ans=0;
int main() {
	cin>>n;
	for(int i=1;i<=n;i++){
		long double x;
		scanf("%Lf",&x),a[i]=round(x*1e9);
		if(a[i]%1000000000==0)ans--;
	}
	for(ll i=0,p2=1;p2<=1000000000000000000ll;i++,p2<<=1){
		for(ll j=0,p5=1;p2*p5<=1000000000000000000ll;j++,p5*=5)b[++b[0]]=p2*p5,A[b[0]]=i,B[b[0]]=j;
	}
	for(ll i=1;i<=n;i++){
		for(ll j=1;j<=b[0];j++){
			if(a[i]%b[j]==0&&a[i]%(b[j]*2)!=0&&a[i]%(b[j]*5)!=0){
				cnt[j]++;//cout<<b[j]<<' '<<A[j]<<' '<<B[j]<<endl;
				break;
			}
		}
	}
	for(int i=1;i<=b[0];i++){
		for(int j=1;j<=b[0];j++)if(A[i]+A[j]>=18&&B[i]+B[j]>=18)ans+=cnt[i]*cnt[j];
	}
	cout<<ans/2<<endl;
	return 0;
}
```