#include<bits/stdc++.h>
using namespace std;
const int maxn=200009;

int n;

int a[maxn],b[maxn];

int f[21][21];
char s[maxn];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%s",s);
		int len=strlen(s);
		int flag=0,cnt=0;long long x=0;
		for(int i=0;i<len;++i){
			if(s[i]=='.'){
				flag=1;continue;
			}
			if(flag)++cnt;
			x=x*10+s[i]-'0';
		}
		while(x%2==0){
			++a[i];x/=2;
		}
		while(x%5==0){
			++b[i];x/=5;
		}
		a[i]+=10-cnt;
		b[i]+=10-cnt;
	}
	long long ans=0;
	for(int i=1;i<=n;++i){
		for(int j=0;j<=20;++j){
			for(int k=0;k<=20;++k){
				if(a[i]+j>=20&&b[i]+k>=20)ans+=f[j][k];
			}
		}
		f[min(a[i],20)][min(b[i],20)]++;
	}
	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>
using namespace std;
const int maxn=200009;

int n;

int a[maxn],b[maxn];

int f[21][21];
char s[maxn];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%s",s);
		int len=strlen(s);
		int flag=0,cnt=0;long long x=0;
		for(int i=0;i<len;++i){
			if(s[i]=='.'){
				flag=1;continue;
			}
			if(flag)++cnt;
			x=x*10+s[i]-'0';
		}
		while(x%2==0){
			++a[i];x/=2;
		}
		while(x%5==0){
			++b[i];x/=5;
		}
		a[i]+=10-cnt;
		b[i]+=10-cnt;
	}
	long long ans=0;
	for(int i=1;i<=n;++i){
		for(int j=0;j<=20;++j){
			for(int k=0;k<=20;++k){
				if(a[i]+j>=20&&b[i]+k>=20)ans+=f[j][k];
			}
		}
		f[min(a[i],20)][min(b[i],20)]++;
	}
	printf("%lld\n",ans);
	return 0;
}
```