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: Wrong Answer)

n = int(input())
a = list(map(int, input().split()))

# [0,border)が負になり、[border,n)が正になるときの操作の回数
def multi_times(border):
    ans = 0
    num = 1
    for i in range(border, n):
        if num <= a[i]:
            num = a[i]
        else:
            tmp = -((-num)//a[i])
            cnt = -(-((tmp-1).bit_length()) // 2)
            num = (4**cnt) * a[i]
            ans += cnt * 2

    num = 1
    for i in range(0, border)[::-1]:
        ans += 1
        if num <= a[i]:
            num = a[i]
        else:
            tmp = -((-num)//a[i])
            cnt = -(-((tmp-1).bit_length()) // 2)
            num = (4**cnt) * a[i]
            ans += cnt * 2
    return ans


ok = n+1
ng = 0
 
while abs(ok-ng) > 1:
    mid = (ok+ng)//2
    if mid == 0 or mid == n:
        print(multi_times(mid))
        exit()
    l = multi_times(mid-1)
    m = multi_times(mid)
    r = multi_times(mid+1)
    if l <= m <= r:
        ok = mid
    elif l >= m >= r:
        ng = mid
    else:
        print(min(l,m,r))
        exit()
mid = (ok+ng)//2
print(min(multi_times(mid),multi_times(max(0, mid-1)),multi_times(min(n, mid+1))))