# include <cstdio>
# define N 200010
char s[20];
int a[N],b[N],f[100][100];
int max(int x,int y) {return x>y?x:y;}
int min(int x,int y) {return x<y?x:y;}
int main() {
	int n,i,j,cnt,maxa=0,maxb=0;
	long long x,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++) {
		scanf("%s",s);
		x=0;
		for(j=0;s[j] && s[j]!='.';j++) x=x*10+s[j]-'0';
		cnt=0;
		if(s[j]) {
			for(j++;s[j];j++,cnt++) x=x*10+s[j]-'0';
		}
		cnt=9-cnt;
		while(cnt--) x*=10;
		while(!(x%2)) a[i]++, x/=2;
		while(!(x%5)) b[i]++, x/=5;
		maxa=max(maxa,a[i]);
		maxb=max(maxb,b[i]);
		f[a[i]][b[i]]++;
	}
	for(i=maxa;i>=0;i--) {
		for(j=maxb;j>=0;j--) f[i][j]+=f[i+1][j]+f[i][j+1]-f[i+1][j+1];
	}
	for(i=1;i<=n;i++) {
		ans+=f[18-a[i]<0?0:18-a[i]][18-b[i]<0?0:18-b[i]];
		if(min(a[i]*2,b[i]*2)>=18) ans--;
	}
	printf("%lld",ans/2);
	return 0;
}
### 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 <cstdio>
# define N 200010
char s[20];
int a[N],b[N],f[100][100];
int max(int x,int y) {return x>y?x:y;}
int min(int x,int y) {return x<y?x:y;}
int main() {
	int n,i,j,cnt,maxa=0,maxb=0;
	long long x,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++) {
		scanf("%s",s);
		x=0;
		for(j=0;s[j] && s[j]!='.';j++) x=x*10+s[j]-'0';
		cnt=0;
		if(s[j]) {
			for(j++;s[j];j++,cnt++) x=x*10+s[j]-'0';
		}
		cnt=9-cnt;
		while(cnt--) x*=10;
		while(!(x%2)) a[i]++, x/=2;
		while(!(x%5)) b[i]++, x/=5;
		maxa=max(maxa,a[i]);
		maxb=max(maxb,b[i]);
		f[a[i]][b[i]]++;
	}
	for(i=maxa;i>=0;i--) {
		for(j=maxb;j>=0;j--) f[i][j]+=f[i+1][j]+f[i][j+1]-f[i+1][j+1];
	}
	for(i=1;i<=n;i++) {
		ans+=f[18-a[i]<0?0:18-a[i]][18-b[i]<0?0:18-b[i]];
		if(min(a[i]*2,b[i]*2)>=18) ans--;
	}
	printf("%lld",ans/2);
	return 0;
}
```