p03198 CADDi 2018 - Negative Doubling_5056
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.

Examples

Input

4
3 1 4 1


Output

3


Input

5
1 2 3 4 5


Output

0


Input

8
657312726 129662684 181537270 324043958 468214806 916875077 825989291 319670097


Output

7
n = int(input())
a = list(map(int,input().split()))
def d4(x,y):
  ret = 0
  while x > y:
    y *= 4
    ret += 1
  return ret
def cost(ls):
  l = len(ls)
  ret = [0]*l
  stack = [(l-1,10**18)]
  for i in range(l-1)[::-1]:
    ret[i] = ret[i+1]
    if ls[i] <= ls[i+1]:
      d = d4(ls[i+1],ls[i])-1
      if d > 0:
        stack.append((i,d))
    else:
      d = d4(ls[i],ls[i+1])
      while d:
        idx,num = stack.pop()
        if d > num:
          ret[i] += (idx-i)*num
          d -= num
        else:
          ret[i] += (idx-i)*d
          num -= d
          d = 0
          if num:
            stack.append((idx,num))
  return ret
cp = cost(a)
cn = cost(a[::-1])
for i in range(n):
  cp[i] *= 2
  cn[i] *= 2
  cn[i] += n-i
ansls = [cp[0],cn[0]]
for i in range(1,n):
  ansls.append(cp[i]+cn[n-i])
print(min(ansls))