#include <bits/stdc++.h>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << #x << " is " << x << endl

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

const int INF = 1e9 + 7;

signed main()
{
	IO_OP;
	
	int n;
	cin >> n;
	map<pi, int> mp;
	ll ans = 0;
	for(int i = 0; i < n; i++) {
		long double x;
		cin >> x;
		ll y = x * 1e9 + 0.5;
		int a = 0, b = 0;
		while(y % 2 == 0) a++, y /= 2;
		while(y % 5 == 0) b++, y /= 5;
		for(auto p:mp) {
			if(p.F.F+a >= 18 && p.F.S+b >= 18) ans += p.S;
		}
		mp[{a, b}]++;
	}
	cout << ans << endl;	
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << #x << " is " << x << endl

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

const int INF = 1e9 + 7;

signed main()
{
	IO_OP;
	
	int n;
	cin >> n;
	map<pi, int> mp;
	ll ans = 0;
	for(int i = 0; i < n; i++) {
		long double x;
		cin >> x;
		ll y = x * 1e9 + 0.5;
		int a = 0, b = 0;
		while(y % 2 == 0) a++, y /= 2;
		while(y % 5 == 0) b++, y /= 5;
		for(auto p:mp) {
			if(p.F.F+a >= 18 && p.F.S+b >= 18) ans += p.S;
		}
		mp[{a, b}]++;
	}
	cout << ans << endl;	
}
```