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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import  java.util.*;

import static java.lang.System.in;

class Main{
    static TreeMap<Long,Integer> tm = new TreeMap<>();
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[] a = new long[n];
        for(int i=0;i<n;i++) a[i]=sc.nextLong();
        long base = 1;
        for(int i=0;i<=31;i++){
            tm.put(base,i);
            base *= i<31?4:1;
        }
        long[] rev = new long[n];
        for(int i=0;i<n;i++) rev[i] = a[n-1-i];
        long[] right = help(a), left = help(rev);
        long ans = 2*right[0];
        for(int i=1;i<n;i++){
            long cur = 2*(right[i]+left[n-i])+i;
            ans = Math.min(ans,cur);
        }
        ans = Math.min(ans,2*left[0]+n);
        System.out.println(ans);

    }
    static long[] help(long[] A){
        int n = A.length;
        long[][] dp = new long[n][16];
        long[] ans = new long[n];
        Arrays.fill(ans,Long.MAX_VALUE);
        ans[n-1]=0;
        for(int j=0;j<16;j++) dp[n-1][j]=j;
        for(int i=n-2;i>=0;i--){
            for(int j=0;j<16;j++){
                long a = A[i]<<(2*j);
                int t = 0;
                while(a>(A[i+1]<<(2*t))) t++;
                if(t<=15) dp[i][j] = j+dp[i+1][t];
                else dp[i][j] = j+dp[i+1][15]+(t-15)*(n-i-1);
                ans [i] = Math.min(ans[i],dp[i][j]);
            }
        }
        return ans;
    }
}