Problem:
Score : 800 points

Problem Statement

There are N positive integers A_1, A_2, ..., A_N.
Takahashi can perform the following operation on these integers any number of times:

Choose 1 \leq i \leq N and multiply the value of A_i by -2.

Notice that he multiplies it by minus two.

He would like to make A_1 \leq A_2 \leq ... \leq A_N holds.
Find the minimum number of operations required. If it is impossible, print -1.

Constraints

1 \leq N \leq 200000

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the answer.

Sample Input 1

4
3 1 4 1

Sample Output 1

3

One possible solution is:

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now 3, 1, 4, -2.

Choose i=1 and multiply the value of A_1 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, -2.

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, 4.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

0

A_1 \leq A_2 \leq ... \leq A_N holds before any operation is performed.

Sample Input 3

8
657312726 129662684 181537270 324043958 468214806 916875077 825989291 319670097

Sample Output 3

7

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;
using ll = long long;

int main(){
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<ll> a(n);
	for(int i = 0; i < n; ++i){ cin >> a[i]; }
	vector<ll> positive(n), negative(n);
	for(int i = n - 2; i >= 0; --i){
		positive[i] = positive[i + 1];
		for(ll j = 1; a[i + 1] * j < a[i]; j *= 4){
			positive[i] += 2 * (n - i - 1);
		}
	}
	if(n >= 2){ negative[1] = 1; }
	for(int i = 2; i < n; ++i){
		negative[i] = negative[i - 1] + 1;
		for(ll j = 1; a[i - 2] * j < a[i - 1]; j *= 4){
			negative[i] += 2 * (i - 1);
		}
	}
	ll answer = 1000000000000000000ll;
	for(int i = 0; i < n; ++i){
		answer = min(answer, negative[i] + positive[i]);
	}
	cout << answer << endl;
	return 0;
}