#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ar array

ll c[19][19];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	int n;
	cin >> n;
	ll ans=0;
	for(int i=0; i<n; ++i) {
		string s;
		cin >> s;
		int p=find(s.begin(), s.end(), '.')-s.begin();
		if(p>=s.size())
			s+='.';
		while(s.size()-p-1<9)
			s+='0';
		ll a=0;
		for(char c : s) {
			if(c=='.')
				continue;
			a=a*10+c-'0';
		}
		int p2=0, p5=0;
		while(p2<18&&a%2==0) {
			++p2;
			a/=2;
		}
		while(p5<18&&a%5==0) {
			++p5;
			a/=5;
		}
		if(p2>=9&&p5>=9)
			--ans;
		++c[p2][p5];
	}
	for(int i1=0; i1<=18; ++i1)
		for(int i2=0; i2<=18; ++i2)
			for(int j1=18-i1; j1<=18; ++j1)
				for(int j2=18-i2; j2<=18; ++j2)
					ans+=c[i1][i2]*c[j1][j2];
	cout << ans/2;
}
### Prompt

Generate 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;

#define ll long long
#define ar array

ll c[19][19];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	int n;
	cin >> n;
	ll ans=0;
	for(int i=0; i<n; ++i) {
		string s;
		cin >> s;
		int p=find(s.begin(), s.end(), '.')-s.begin();
		if(p>=s.size())
			s+='.';
		while(s.size()-p-1<9)
			s+='0';
		ll a=0;
		for(char c : s) {
			if(c=='.')
				continue;
			a=a*10+c-'0';
		}
		int p2=0, p5=0;
		while(p2<18&&a%2==0) {
			++p2;
			a/=2;
		}
		while(p5<18&&a%5==0) {
			++p5;
			a/=5;
		}
		if(p2>=9&&p5>=9)
			--ans;
		++c[p2][p5];
	}
	for(int i1=0; i1<=18; ++i1)
		for(int i2=0; i2<=18; ++i2)
			for(int j1=18-i1; j1<=18; ++j1)
				for(int j2=18-i2; j2<=18; ++j2)
					ans+=c[i1][i2]*c[j1][j2];
	cout << ans/2;
}
```