#include <bits/stdc++.h>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<lint, lint> pi;
const int MAXN = 530000;
const int mod = 200003;

map<pi, int> mp;

int main(){
	int n; scanf("%d",&n);
	lint ans = 0;
	for(int i=0; i<n; i++){
		double x; scanf("%lf",&x);
		lint q = (lint)round(x * 1e9);
		if(q % lint(1e9) == 0) ans--;
		int c0 = -9, c1 = -9;
		while(q % 2 == 0){
			q /= 2;
			c0++;
		}
		while(q % 5 == 0){
			q /= 5;
			c1++;
		}
		mp[pi(c0, c1)]++;
	}
	for(auto &[x1, y1] : mp){
		for(auto &[x2, y2] : mp){
			if(x1.first + x2.first >= 0 && x1.second + x2.second >= 0) ans += 1ll * y1 * y2;
		}
	}
	ans /= 2;
	cout << ans << endl;
}

### Prompt

Your challenge is to write 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>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<lint, lint> pi;
const int MAXN = 530000;
const int mod = 200003;

map<pi, int> mp;

int main(){
	int n; scanf("%d",&n);
	lint ans = 0;
	for(int i=0; i<n; i++){
		double x; scanf("%lf",&x);
		lint q = (lint)round(x * 1e9);
		if(q % lint(1e9) == 0) ans--;
		int c0 = -9, c1 = -9;
		while(q % 2 == 0){
			q /= 2;
			c0++;
		}
		while(q % 5 == 0){
			q /= 5;
			c1++;
		}
		mp[pi(c0, c1)]++;
	}
	for(auto &[x1, y1] : mp){
		for(auto &[x2, y2] : mp){
			if(x1.first + x2.first >= 0 && x1.second + x2.second >= 0) ans += 1ll * y1 * y2;
		}
	}
	ans /= 2;
	cout << ans << endl;
}

```