#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define int long long
#define m_k make_pair
using namespace std;
const int N=2*1e5+100;
int n,a[N],b[N],sum[100][100],ans;
char s[1000];
signed main()
{
	scanf("%lld",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",s);
		int len=strlen(s);
		int x=0,cnt=0;
		bool bl=1;
		for (int j=0;j<len;j++)
		{
			if (s[j]=='.')
			{
				bl=0;
				continue;
			}
			if (!bl) cnt++;
			x=x*10ll+s[j]-'0';
		}
		for (int j=1;j<=9-cnt;j++) x*=10ll;
		while (x%2==0) a[i]++,x/=2ll;
		while (x%5==0) b[i]++,x/=5ll;
		sum[a[i]][b[i]]++;
		if (a[i]+a[i]>=18 && b[i]+b[i]>=18) ans--;
	}
	for (int i=70;i>=0;i--)
	{
		for (int j=70;j>=0;j--) sum[i][j]+=sum[i][j+1];
	}
	for (int j=70;j>=0;j--)
	{
		for (int i=70;i>=0;i--) sum[i][j]+=sum[i+1][j];
	}
	for (int i=1;i<=n;i++) ans+=sum[max(18-a[i],0ll)][max(18-b[i],0ll)];
	ans/=2;
	printf("%lld\n",ans);
}
### Prompt

Develop a solution in cpp to the problem described below:

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 inf 0x3f3f3f3f
#define int long long
#define m_k make_pair
using namespace std;
const int N=2*1e5+100;
int n,a[N],b[N],sum[100][100],ans;
char s[1000];
signed main()
{
	scanf("%lld",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",s);
		int len=strlen(s);
		int x=0,cnt=0;
		bool bl=1;
		for (int j=0;j<len;j++)
		{
			if (s[j]=='.')
			{
				bl=0;
				continue;
			}
			if (!bl) cnt++;
			x=x*10ll+s[j]-'0';
		}
		for (int j=1;j<=9-cnt;j++) x*=10ll;
		while (x%2==0) a[i]++,x/=2ll;
		while (x%5==0) b[i]++,x/=5ll;
		sum[a[i]][b[i]]++;
		if (a[i]+a[i]>=18 && b[i]+b[i]>=18) ans--;
	}
	for (int i=70;i>=0;i--)
	{
		for (int j=70;j>=0;j--) sum[i][j]+=sum[i][j+1];
	}
	for (int j=70;j>=0;j--)
	{
		for (int i=70;i>=0;i--) sum[i][j]+=sum[i+1][j];
	}
	for (int i=1;i<=n;i++) ans+=sum[max(18-a[i],0ll)][max(18-b[i],0ll)];
	ans/=2;
	printf("%lld\n",ans);
}
```