#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn=200005,maxm=70;
struct datas { int A,B; } cow[maxn]; 
int n,cnt[maxm][maxm];
LL ans,num[maxn];
string S;
int main()
{
	scanf("%d",&n);
	for (int i=1,A,B;i<=n;i++)
	{
		cin>>S;
		int len=S.size(),pos=S.find('.');
		if (pos==string::npos) pos=len-1;
		for (int j=0,J=len+9-(len-pos-1);j<J;j++)
		{
			if (j<len && S[j]!='.') num[i]=num[i]*10+S[j]-48;
			if (j>=len) num[i]*=10;
		}
		//cerr<<">>>>>>"<<num[i]<<endl;
		A=B=0;
		while (!(num[i]&1))
		{
			A++;
			num[i]>>=1;
		}
		while (!(num[i]%5))
		{
			B++;
			num[i]/=5;
		}
		for (int x=max(18-A,0);x<maxm;x++)
			for (int y=max(18-B,0);y<maxm;y++)
				ans+=cnt[x][y];
		cnt[A][B]++;
	}
	printf("%lld\n",ans);
	return 0;
}

### Prompt

In CPP, your task is to solve 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>
#define LL long long
using namespace std;
const int maxn=200005,maxm=70;
struct datas { int A,B; } cow[maxn]; 
int n,cnt[maxm][maxm];
LL ans,num[maxn];
string S;
int main()
{
	scanf("%d",&n);
	for (int i=1,A,B;i<=n;i++)
	{
		cin>>S;
		int len=S.size(),pos=S.find('.');
		if (pos==string::npos) pos=len-1;
		for (int j=0,J=len+9-(len-pos-1);j<J;j++)
		{
			if (j<len && S[j]!='.') num[i]=num[i]*10+S[j]-48;
			if (j>=len) num[i]*=10;
		}
		//cerr<<">>>>>>"<<num[i]<<endl;
		A=B=0;
		while (!(num[i]&1))
		{
			A++;
			num[i]>>=1;
		}
		while (!(num[i]%5))
		{
			B++;
			num[i]/=5;
		}
		for (int x=max(18-A,0);x<maxm;x++)
			for (int y=max(18-B,0);y<maxm;y++)
				ans+=cnt[x][y];
		cnt[A][B]++;
	}
	printf("%lld\n",ans);
	return 0;
}

```