Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Constraints

* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_0 A_1 A_2 \cdots A_N


Output

Print the answer as an integer.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
n=int(input())
l=list(map(int,input().split()))
xl=list(reversed(l))
g=[]
a=1
k=0
f=0
for i in xl:
  k+=i
  g.append(k)
g.reverse()
for i in range(len(g)):
  if g[i]>=a:
    g[i]=a
  a-=l[i]
  a*=2
  if a<=0 and i!=len(g)-1:
    f+=1
    break
  if a<0 and i==len(g)-1:
    f+=1
    break
if f==0 and n!=0:
  print(sum(g))
elif n==0:
  if l[0]==1:
    print(1)
  else:
    print(-1)
else:
  print(-1)