#include <bits/stdc++.h>

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

using namespace std;
using lint=long long;

int main(){
	int n; cin>>n;
	vector<lint> a(n);
	rep(i,n){
		string s; cin>>s;
		int j=s.find('.');
		if(j==string::npos){
			lint x=stoll(s);
			a[i]=x*lint(1e9);
		}
		else{
			lint x=stoll(s.substr(0,j));
			string t=s.substr(j+1);
			while(t.length()<9) t+='0';
			lint y=stoll(t);
			a[i]=x*lint(1e9)+y;
		}
	}

	map<pair<int,int>,lint> f;
	for(lint x:a){
		int c2=0,c5=0;
		while(x%2==0) x/=2, c2++;
		while(x%5==0) x/=5, c5++;
		++f[{c2,c5}];
	}

	lint ans=0;
	for(auto p:f) for(auto q:f) {
		int c2=p.first.first +q.first.first;
		int c5=p.first.second+q.first.second;
		if(min(c2,c5)>=18) ans+=p.second*q.second;
	}
	rep(i,n) if(a[i]%lint(1e9)==0) ans--;
	ans/=2;
	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 rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

int main(){
	int n; cin>>n;
	vector<lint> a(n);
	rep(i,n){
		string s; cin>>s;
		int j=s.find('.');
		if(j==string::npos){
			lint x=stoll(s);
			a[i]=x*lint(1e9);
		}
		else{
			lint x=stoll(s.substr(0,j));
			string t=s.substr(j+1);
			while(t.length()<9) t+='0';
			lint y=stoll(t);
			a[i]=x*lint(1e9)+y;
		}
	}

	map<pair<int,int>,lint> f;
	for(lint x:a){
		int c2=0,c5=0;
		while(x%2==0) x/=2, c2++;
		while(x%5==0) x/=5, c5++;
		++f[{c2,c5}];
	}

	lint ans=0;
	for(auto p:f) for(auto q:f) {
		int c2=p.first.first +q.first.first;
		int c5=p.first.second+q.first.second;
		if(min(c2,c5)>=18) ans+=p.second*q.second;
	}
	rep(i,n) if(a[i]%lint(1e9)==0) ans--;
	ans/=2;
	printf("%lld\n",ans);

	return 0;
}

```