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)

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 construct_dp(a: &Vec<u64>) -> Vec<Vec<usize>> {
    let mut dp = vec![vec![0; 16]; a.len()];
    for i in 0..16 {
        dp[a.len() - 1][i] = i;
    }

    for i in (1..a.len()).rev() {
        for j in 0..16 {
            dp[i - 1][j] = j;

            let a1 = a[i - 1] * 4u64.pow(j as u32);
            let mut a2 = a[i];
            let mut inc: usize = 0;
            while a2 < a1 {
                inc += 1;
                a2 *= 4;
            }
            if inc > 15 {
                dp[i - 1][j] += (inc - 15) * (a.len() - 1 - i);
                inc = 15;
            }
            dp[i - 1][j] += dp[i][inc];
        }
    }
    dp
}

fn main() {
    let n: usize = read();
    let mut a = read_vec::<u64>();
    let dp_forward = construct_dp(&a);

    a.reverse();
    let dp_backward = construct_dp(&a);
    let mut ans = std::cmp::min(dp_forward[0][0] * 2, dp_backward[0][0] * 2 + a.len());

    for i in 1..n-1 {
        // println!("{:?}", dp_forward[i]);
        let tmp_ans = dp_forward[i][0] * 2 + i + dp_backward[n - i][0] * 2;
        // println!("i {0} dp_forward[i][0] {1} dp_backward[n - i - 1][0] {2} {3}",
        //          i, dp_forward[i][0] * 2, dp_backward[n - i - 1][0] * 2, tmp_ans);
        ans = std::cmp::min(ans, tmp_ans);
    }

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