#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
const ll N = 1e9+7;


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	ll t = 1;
	// cin>>t;

	while(t--){
		ll n;
		cin>>n;

		map<pair<ll,ll>,ll>m;
		double a;
		for(ll i=0;i<n;++i){
			cin>>a;
			ll x = llround(a * 1e9);

			ll two = 0, five = 0;
			while(x % 2 == 0){
				++two;
				x /= 2;
			}
			while(x % 5 == 0){
				++five;
				x /= 5;
			}

			m[{two,five}]++;

		}

		ll ans = 0;
		for(auto &x : m){
			for(auto &y : m){
				if(x.first.first + y.first.first >= 18 and x.first.second + y.first.second >= 18){
					if(x == y) ans += x.second * ( x.second -1 ) / 2;
					else if(x < y) ans += x.second * y.second;
				}
			}
		}

		cout<<ans<<"\n";

	}

	return 0;

}

### Prompt

Please create a solution in cpp 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;
typedef long long int ll;
const ll N = 1e9+7;


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	ll t = 1;
	// cin>>t;

	while(t--){
		ll n;
		cin>>n;

		map<pair<ll,ll>,ll>m;
		double a;
		for(ll i=0;i<n;++i){
			cin>>a;
			ll x = llround(a * 1e9);

			ll two = 0, five = 0;
			while(x % 2 == 0){
				++two;
				x /= 2;
			}
			while(x % 5 == 0){
				++five;
				x /= 5;
			}

			m[{two,five}]++;

		}

		ll ans = 0;
		for(auto &x : m){
			for(auto &y : m){
				if(x.first.first + y.first.first >= 18 and x.first.second + y.first.second >= 18){
					if(x == y) ans += x.second * ( x.second -1 ) / 2;
					else if(x < y) ans += x.second * y.second;
				}
			}
		}

		cout<<ans<<"\n";

	}

	return 0;

}

```