#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// make it understandable one day...
const int N=2e5+1;
int n;
int a[N],b[N];
int f[31][31];
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n;
	for(int i=1; i<=n ;i++){
		string s;cin >> s;
		ll x=0;
		bool d=false;
		int g=0;
		for(auto c:s){
			if(c=='.'){
				d=true;continue;
			}
			x=x*10+c-48;
			if(d) g++;
		}
		while(g++<9) x=x*10;
		while(x%2==0){
			a[i]++;x/=2;
		}
		while(x%5==0){
			b[i]++;x/=5;
		}
		a[i]=min(a[i],18);
		b[i]=min(b[i],18);
	}
	for(int i=1; i<=n ;i++){
		f[a[i]][b[i]]++;
	}
	for(int i=18; i>=0 ;i--){
		for(int j=18; j>=0 ;j--){
			f[i][j]+=f[i+1][j]+f[i][j+1]-f[i+1][j+1];
		}
	}
	ll ans=0;
	for(int i=1; i<=n ;i++){
		ans+=f[18-a[i]][18-b[i]];
		if(a[i]*2>=18 && b[i]*2>=18) ans--;
	}
	ans/=2;
	cout << ans << '\n';
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
// make it understandable one day...
const int N=2e5+1;
int n;
int a[N],b[N];
int f[31][31];
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n;
	for(int i=1; i<=n ;i++){
		string s;cin >> s;
		ll x=0;
		bool d=false;
		int g=0;
		for(auto c:s){
			if(c=='.'){
				d=true;continue;
			}
			x=x*10+c-48;
			if(d) g++;
		}
		while(g++<9) x=x*10;
		while(x%2==0){
			a[i]++;x/=2;
		}
		while(x%5==0){
			b[i]++;x/=5;
		}
		a[i]=min(a[i],18);
		b[i]=min(b[i],18);
	}
	for(int i=1; i<=n ;i++){
		f[a[i]][b[i]]++;
	}
	for(int i=18; i>=0 ;i--){
		for(int j=18; j>=0 ;j--){
			f[i][j]+=f[i+1][j]+f[i][j+1]-f[i+1][j+1];
		}
	}
	ll ans=0;
	for(int i=1; i<=n ;i++){
		ans+=f[18-a[i]][18-b[i]];
		if(a[i]*2>=18 && b[i]*2>=18) ans--;
	}
	ans/=2;
	cout << ans << '\n';
}
```