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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import numpy as np


def calc_mul4_dp(n, aaa):
    dp = np.array([list(range(16))] * (n + 1), dtype=np.int64)
    arg = np.arange(1, n + 1, dtype=np.int64)

    next_a = aaa[-1]
    for i in range(n - 2, -1, -1):
        a = aaa[i]
        if a > next_a:
            cnt = 0  # a[i]以上になるためにa[i+1]に4をかけなければならない最小回数
            b = next_a
            while a > b:
                b <<= 2
                cnt += 1
            dp[i, :-cnt] += dp[i + 1, cnt:]
            dp[i, -cnt:] += dp[i + 1, -1] + arg[:cnt] * (n - i - 1)
        else:
            cnt = 0  # a[i]に4をかけても次に影響が出ない最大回数
            b = a
            while next_a >= b:
                b <<= 2
                cnt += 1
            cnt -= 1
            dp[i, :cnt] += dp[i + 1, 0]
            dp[i, cnt:] += dp[i + 1, :16 - cnt]
        next_a = a
    return dp


def solve(n, aaa):
    fdp = calc_mul4_dp(n, aaa)
    bdp = calc_mul4_dp(n, aaa[::-1])[::-1]
    mul4dp = (fdp.T[0] + bdp.T[0]) * 2
    return (mul4dp + np.arange(n + 1)).min()


def test():
    import random
    n = 200000
    aaa = random.choices(range(1, 10 ** 9), k=n)
    print(n)
    print(aaa)
    print(solve(n, aaa))


n = int(input())
aaa = list(map(int, input().split()))
print(solve(n, aaa))