#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=1e9; 
int a[10];
int n;
string s;
ll num1,num2;
void go(){
	num1=0,num2=0;
	int cur=0;
	while(cur<(int)s.size()&&s[cur]!='.')num1=num1*10+s[cur]-'0',cur++;
	if((int)s.size()==cur)return;
	cur++;
	int now=8;
	while(cur<(int)s.size())num2+=a[now]*(s[cur]-'0'),now--,cur++;
}
int wants[20][20];
int p2,p5;
void intbreak(ll x){
	p2=0,p5=0;
	while(x%2==0)p2++,x/=2;
	while(x%5==0)p5++,x/=5;
}
ll ans=0;
int main(){
	cin>>n;
	a[0]=1;
	for(int i=1;i<9;i++)a[i]=a[i-1]*10;
	for(int i=1;i<=n;i++){
		cin>>s;
		go();
		num1=num1*INF+num2;
		intbreak(num1);
		p2=min(p2,18),p5=min(p5,18);
		ans+=wants[p2][p5];
		int P2=18-p2,P5=18-p5;
		for(int j=P2;j<19;j++)for(int k=P5;k<19;k++)wants[j][k]++;
	}
	cout<<ans;
}
### Prompt

Please create a solution in CPP 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 ll;
const ll INF=1e9; 
int a[10];
int n;
string s;
ll num1,num2;
void go(){
	num1=0,num2=0;
	int cur=0;
	while(cur<(int)s.size()&&s[cur]!='.')num1=num1*10+s[cur]-'0',cur++;
	if((int)s.size()==cur)return;
	cur++;
	int now=8;
	while(cur<(int)s.size())num2+=a[now]*(s[cur]-'0'),now--,cur++;
}
int wants[20][20];
int p2,p5;
void intbreak(ll x){
	p2=0,p5=0;
	while(x%2==0)p2++,x/=2;
	while(x%5==0)p5++,x/=5;
}
ll ans=0;
int main(){
	cin>>n;
	a[0]=1;
	for(int i=1;i<9;i++)a[i]=a[i-1]*10;
	for(int i=1;i<=n;i++){
		cin>>s;
		go();
		num1=num1*INF+num2;
		intbreak(num1);
		p2=min(p2,18),p5=min(p5,18);
		ans+=wants[p2][p5];
		int P2=18-p2,P5=18-p5;
		for(int j=P2;j<19;j++)for(int k=P5;k<19;k++)wants[j][k]++;
	}
	cout<<ans;
}
```