#include<cstdio>
#include<cmath>
#include<algorithm>
#define lf double
#define ll long long
#define N 200005

int n;

int cnt[50][50],cnt2[N],cnt5[N];

ll ans;

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		lf A;
		scanf("%lf",&A);
		ll a=(llround)(A*1e9);
		while(a%2==0){
			cnt2[i]++;
			a/=2;
		}
		while(a%5==0){
			cnt5[i]++;
			a/=5;
		}
		cnt[cnt2[i]][cnt5[i]]++;
		if(cnt2[i]>=9&&cnt5[i]>=9)
			ans--;
		cnt2[i]=std::max(18-cnt2[i],0);
		cnt5[i]=std::max(18-cnt5[i],0);
	}
	for(int i=48;i>=0;i--)
		for(int j=48;j>=0;j--)
			cnt[i][j]+=cnt[i+1][j]+cnt[i][j+1]-cnt[i+1][j+1];
	for(int i=1;i<=n;i++)
		ans+=cnt[cnt2[i]][cnt5[i]];
	printf("%lld\n",ans/2);
} 
### 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<cstdio>
#include<cmath>
#include<algorithm>
#define lf double
#define ll long long
#define N 200005

int n;

int cnt[50][50],cnt2[N],cnt5[N];

ll ans;

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		lf A;
		scanf("%lf",&A);
		ll a=(llround)(A*1e9);
		while(a%2==0){
			cnt2[i]++;
			a/=2;
		}
		while(a%5==0){
			cnt5[i]++;
			a/=5;
		}
		cnt[cnt2[i]][cnt5[i]]++;
		if(cnt2[i]>=9&&cnt5[i]>=9)
			ans--;
		cnt2[i]=std::max(18-cnt2[i],0);
		cnt5[i]=std::max(18-cnt5[i],0);
	}
	for(int i=48;i>=0;i--)
		for(int j=48;j>=0;j--)
			cnt[i][j]+=cnt[i+1][j]+cnt[i][j+1]-cnt[i+1][j+1];
	for(int i=1;i<=n;i++)
		ans+=cnt[cnt2[i]][cnt5[i]];
	printf("%lld\n",ans/2);
} 
```