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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define NOT_ZERO 1
#define MAX(a,b) (a>b?a:b)

int main()
{
    long long n, i, a[200001]={NOT_ZERO};
    long long count = 0;
    scanf("%lld",&n);
    for(i=1; i<=n; i++) {
        scanf("%lld",&a[i]);
    }
    
    i = 1;
    if(a[i]>a[i+1]) {
        a[i] *= (-2);
        count += 1;
    }

    for(i=2; i<=n-1; i++) {
        if(a[i]>a[i+1]) {
            if(a[i-1]<=a[i]*(-2)) {
                a[i] *= (-2);
                count += 1;
            } else {
                while(a[i]>a[i+1]) {
                    a[i+1] *= 4;
                    count += 2;
                }
            }
        }
    }

    printf("%lld",count);
    return 0;
}