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: Accepted)

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

const long long INF=1LL<<61;

int main(){
	int n; scanf("%d",&n);
	vector<lint> a(n);
	rep(i,n) scanf("%lld",&a[i]);

	vector<lint> sum1(n+1);
	{
		auto tmp=a;

		vector<int> right(n);
		rep(i,n) right[i]=i;

		for(int i=n-2;i>=0;i--){
			sum1[i]=sum1[i+1];
			int pos=i;
			while(pos<n-1 && a[pos]>a[pos+1]){
				int r=right[pos+1];
				while(a[pos]>a[pos+1]){
					sum1[i]+=2*(r-pos);
					a[pos+1]*=4;
					if(pos+1<r) a[r]*=4;
				}
				pos=r;
			}
			right[i]=pos;
		}

		a=tmp;
	}

	vector<lint> sum2(n+1);
	{
		auto tmp=a;

		vector<int> left(n);
		rep(i,n) left[i]=i;

		for(int i=2;i<=n;i++){
			sum2[i]=sum2[i-1];

			int pos=i-1;
			while(pos>=1 && a[pos-1]<a[pos]){
				int l=left[pos-1];
				while(a[pos-1]<a[pos]){
					sum2[i]+=2*(pos-l);
					a[pos-1]*=4;
					if(l<pos-1) a[l]*=4;
				}
				pos=l;
			}
			left[i-1]=pos;
		}

		a=tmp;
	}

	lint ans=INF;
	rep(i,n+1) ans=min(ans,sum1[i]+sum2[i]+i);
	printf("%lld\n",ans);

	return 0;
}