#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 1e18;
#define REP(i, n) for(int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int main() {
	int n; cin >> n;
	map<P,int> mp;
	vector<P> p(n);
	double a;
	REP(i,n){
		cin >> a;
		ll b=(a*1000000000+0.1);
		int d2=0, d5=0;
		while(b%2==0){
			b/=2;
			d2++;
		}
		while(b%5==0){
			b/=5;
			d5++;
		}
		d2=min(18,d2);
		d5=min(18,d5);
		mp[{d2,d5}]++;
		p[i]={d2,d5};
	}
	ll ans=0;
	REP(i,n){
		for(int j=max(0ll,18-p[i].first);j<=18;j++){
			for(int k=max(0ll,18-p[i].second);k<=18;k++){
				ans+=mp[{j,k}];
				if(p[i].first==j&&p[i].second==k)
					ans--;
			}
		}
	}
	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 int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 1e18;
#define REP(i, n) for(int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int main() {
	int n; cin >> n;
	map<P,int> mp;
	vector<P> p(n);
	double a;
	REP(i,n){
		cin >> a;
		ll b=(a*1000000000+0.1);
		int d2=0, d5=0;
		while(b%2==0){
			b/=2;
			d2++;
		}
		while(b%5==0){
			b/=5;
			d5++;
		}
		d2=min(18,d2);
		d5=min(18,d5);
		mp[{d2,d5}]++;
		p[i]={d2,d5};
	}
	ll ans=0;
	REP(i,n){
		for(int j=max(0ll,18-p[i].first);j<=18;j++){
			for(int k=max(0ll,18-p[i].second);k<=18;k++){
				ans+=mp[{j,k}];
				if(p[i].first==j&&p[i].second==k)
					ans--;
			}
		}
	}
	cout << ans/2 << endl;
	return 0;
}
```