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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from math import ceil,log
N = int(input())
A = list(map(int,input().split(" ")))
A2 = list(map(lambda x: x*(-2),A))
dp1 = [ [0]*16 for _ in range(N) ]
dp2 = [ [0]*16 for _ in range(N) ]
pow4 = [ pow(4,x) for x in range(16) ]

for x in range(16):
    dp1[-1][x] = x
    dp2[0][x] = x
for l in range(N-2,-1,-1):
    for x in range(16):
        if pow4[x]*A[l] <= A[l+1]:
            dp1[l][x] = dp1[l+1][0] + x
        else:
            k = ceil(log(pow4[x]*A[l]/A[l+1],4))
            if k > 15:
                dp1[l][x] = dp1[l+1][15] + (N-l+1)*(k-15)
            else:
                dp1[l][x] = dp1[l+1][k] + x

for l in range(1,N):
    for x in range(16):
        if pow4[x]*A2[l] >= A2[l-1]:
            dp2[l][x] = dp2[l-1][0] + x
        else:
            k = ceil(log(pow4[x]*A2[l]/A2[l-1],4))
            if k > 15:
                dp2[l][x] = dp2[l-1][15] + (l+1)*(k -15)
            else:
                dp2[l][x] = dp2[l-1][k] + x


ans = min(2*dp1[0][0],2*dp2[N-1][0]+N)
for i in range(1,N-1):
    ans = min(ans,i+2*(dp1[i][0]+dp2[i-1][0]))

print(ans)