#include <bits/stdc++.h>
#define ll long long 
#define ld long double
using namespace std;
const ll INF = 1e18;
const ll MOD = 998244353;



int32_t main() {
	ios::sync_with_stdio(!cin.tie(NULL));

	int n;
	cin >> n;

	double arr[n];
	map<pair<int, int>, ll> mp;

	for(int i = 0; i < n; i++) {
		cin >> arr[i];
		arr[i] *= 1e9;
		ll tmp = llround(arr[i]);
		int a = 0, b = 0;
		while(tmp % 2ll == 0ll)		tmp /= 2ll, a++;
		while(tmp % 5ll == 0ll)		tmp /= 5ll, b++;
		mp[make_pair(a, b)]++;
	}

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

	cout << ans;
}


### 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 <bits/stdc++.h>
#define ll long long 
#define ld long double
using namespace std;
const ll INF = 1e18;
const ll MOD = 998244353;



int32_t main() {
	ios::sync_with_stdio(!cin.tie(NULL));

	int n;
	cin >> n;

	double arr[n];
	map<pair<int, int>, ll> mp;

	for(int i = 0; i < n; i++) {
		cin >> arr[i];
		arr[i] *= 1e9;
		ll tmp = llround(arr[i]);
		int a = 0, b = 0;
		while(tmp % 2ll == 0ll)		tmp /= 2ll, a++;
		while(tmp % 5ll == 0ll)		tmp /= 5ll, b++;
		mp[make_pair(a, b)]++;
	}

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

	cout << ans;
}


```