#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int mxN=2e5;

int main() {
//#ifndef ONLINE_JUGDE
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
//#endif

    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int n;
    cin>>n;
    map<pair<int,int>,int> cnt;
    for(int i=0;i<n;i++){
    	double num;
    	cin>>num;
    	ll x=llround(num*1e9);
    	int two=0,five=0;
    	while(x%2==0){
    		two++;
    		x/=2;
		}
		while(x%5==0){
			five++;
			x/=5;
		}
		cnt[{two,five}]++;
	}
    ll ans=0;
    for(auto i:cnt){
    	for(auto j:cnt){
    		if(i.first.first+j.first.first>17&&i.first.second+j.first.second>17){
    			if(i<j)
    				ans+=(ll)i.second*j.second;
    			else if(i==j)
    				ans+=(ll)i.second*(j.second-1)/2;
			}
		}
	}
	cout<<ans;
    return 0;
}
### Prompt

Generate 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;
#define ll long long
const int mxN=2e5;

int main() {
//#ifndef ONLINE_JUGDE
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
//#endif

    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int n;
    cin>>n;
    map<pair<int,int>,int> cnt;
    for(int i=0;i<n;i++){
    	double num;
    	cin>>num;
    	ll x=llround(num*1e9);
    	int two=0,five=0;
    	while(x%2==0){
    		two++;
    		x/=2;
		}
		while(x%5==0){
			five++;
			x/=5;
		}
		cnt[{two,five}]++;
	}
    ll ans=0;
    for(auto i:cnt){
    	for(auto j:cnt){
    		if(i.first.first+j.first.first>17&&i.first.second+j.first.second>17){
    			if(i<j)
    				ans+=(ll)i.second*j.second;
    			else if(i==j)
    				ans+=(ll)i.second*(j.second-1)/2;
			}
		}
	}
	cout<<ans;
    return 0;
}
```