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: Time Limit Exceeded)


import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Main m = new Main();
		m.run();
	}

	Scanner sc = new Scanner(System.in);

	void run() {
		int n  =sc.nextInt();
		long a[] = new long[n];
		
		for(int i =0 ;i < n ; i++){
			a[i] = sc.nextInt();
		}
		
		long ans = Long.MAX_VALUE;
		for(int i = 0;  i< n ; i++){
			long sum = i;
			sum += count(i,0,a)*2;
			sum += countRev(i,0,a)*2;
//			System.out.println( i);
//			System.out.println( count(i,0,a)*2);
//			System.out.println( countRev(i,0,a)*2);
			ans = Math.min(sum, ans);
		}
		System.out.println(ans);
	}
	
	Map<String,Long> map = new HashMap<String,Long>();
	Map<String,Long> mapRev = new HashMap<String,Long>();
	
	long count(int i, long b, long[] a){
		if(i >= a.length ){
			return 0;
		}

		String key = i+"_"+b;
		if(map.containsKey(key)){
			return map.get(key);
		}
		
		long res = 0;
		long x = 0;
		long now = a[i] << (b*2L);
		
		if(i+1 < a.length){
			for(;now > (a[i+1] << (x*2L));x++){}
			res = count(i+1,x, a) + x;
		}
		
		map.put(key, res);
		return res;
	}

	long countRev(int i, long b, long[] a){
		if(i < 0){
			return 0;
		}

		String key = i+"_"+b;
		if(mapRev.containsKey(key)){
			return mapRev.get(key);
		}
		
		long res = 0;
		
		long x = 0;
		long now = a[i] << (b*2);
		
		if(i - 1 >= 0){
			for(;now > (a[i-1] << (x*2L));x++){}
			res = countRev(i-1,x, a) + x;
		}
		
		mapRev.put(key, res);
		return res;
	}

	
	
}