#include<bits/stdc++.h>
#define ll long long 
#define rint register int 
#define pa pair<ll,ll>
#define mk make_pair
#define db double
using namespace std;
int N;
map<pa,int> mp;
int main(){
	cin>>N;
	for(rint i=1;i<=N;++i){
		long db x;
		cin>>x;
		ll t; 
		t=llround(x*1e9);
		int cnt1=0,cnt2=0;
		while(t%2==0) {
			cnt1++;
			t/=2;
		}
		while(t%5==0) {
			cnt2++;
			t/=5;
		}
		mp[mk(cnt1,cnt2)]++;
	}
	ll ans=0;
	for(auto i:mp){
		for(auto j:mp) {
			if(i==j) {
				if(i.first.first*2>=18&&i.first.second*2>=18)
					ans+=1ll*i.second*(i.second-1);
			}
			else {
				if(i.first.first+j.first.first>=18&&i.first.second+j.first.second>=18)
				ans+=1ll*i.second*j.second;
			}
		}
	}cout<<ans/2;
	return 0;
}
### 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>
#define ll long long 
#define rint register int 
#define pa pair<ll,ll>
#define mk make_pair
#define db double
using namespace std;
int N;
map<pa,int> mp;
int main(){
	cin>>N;
	for(rint i=1;i<=N;++i){
		long db x;
		cin>>x;
		ll t; 
		t=llround(x*1e9);
		int cnt1=0,cnt2=0;
		while(t%2==0) {
			cnt1++;
			t/=2;
		}
		while(t%5==0) {
			cnt2++;
			t/=5;
		}
		mp[mk(cnt1,cnt2)]++;
	}
	ll ans=0;
	for(auto i:mp){
		for(auto j:mp) {
			if(i==j) {
				if(i.first.first*2>=18&&i.first.second*2>=18)
					ans+=1ll*i.second*(i.second-1);
			}
			else {
				if(i.first.first+j.first.first>=18&&i.first.second+j.first.second>=18)
				ans+=1ll*i.second*j.second;
			}
		}
	}cout<<ans/2;
	return 0;
}
```