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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::collections::HashMap;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

fn count(a: &Vec<f64>) -> Vec<i64> {
    let mut last_num = a[0];
    let mut count = vec![0i64; a.len()];
    for i in 1..a.len() {
        let mut num = a[i];
        if num < last_num {
            let mut temp_inc = (last_num - num).ceil() as i64;
            if temp_inc % 2 == 1 {
                temp_inc += 1;
            }
            count[i] = temp_inc;
            num += temp_inc as f64;
        } else {
            count[i] = 0;
        }
        last_num = num;
    }
    count
}

fn main() {
    let n: usize = read();
    let a = read_vec::<i64>();
    let mut a = a.iter().map(|x| (*x as f64).log2()).collect::<Vec<_>>();
    // println!("{:?}", a);

    let ans1 = count(&a);
    a.reverse();
    let ans2 = count(&a);
    let mut sum_ans1 = vec![0i64; n];
    sum_ans1[0] = ans1[0];
    for i in 1..n {
        sum_ans1[i] = sum_ans1[i - 1] + ans1[i];
    }
    let mut sum_sum_ans1 = vec![0i64; n];
    sum_sum_ans1[0] = sum_ans1[0];
    for i in 1..n {
        sum_sum_ans1[i] = sum_sum_ans1[i - 1] + sum_ans1[i];
    }
    // println!("{:?}", sum_sum_ans1);
    let mut sum_ans2 = vec![0i64; n];
    sum_ans2[0] = ans2[0];
    for i in 1..n {
        sum_ans2[i] = sum_ans2[i - 1] + ans2[i];
    }
    let mut sum_sum_ans2 = vec![0i64; n];
    sum_sum_ans2[0] = sum_ans2[0];
    for i in 1..n {
        sum_sum_ans2[i] = sum_sum_ans2[i - 1] + sum_ans2[i];
    }
    // println!("{:?}", sum_sum_ans2);
    let mut ans = std::cmp::min(sum_sum_ans1[n - 1], sum_sum_ans2[n - 1]);
    for i in 1..n - 1 {
        let mut temp_ans = sum_sum_ans1[n - 1] - sum_sum_ans1[i] * (n - i) as i64;
        // println!("{:?}", (sum_sum_ans1[n - 1], sum_sum_ans1[i], n - i));
        temp_ans += sum_ans2[n - 1] - sum_sum_ans2[n - i] * (i) as i64;
        // println!("{:?}", (sum_sum_ans2[n - 1], sum_sum_ans2[n - i], i));
        temp_ans += i as i64;
        // println!("temp_ans {:?}", temp_ans);
        ans = std::cmp::min(ans, temp_ans);
    }

    println!("{:?}", ans);
}