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())
b0=input().split()
b=[int(x) for x in b0]

ret=0

if b[-1]==0:
  ret=-1
else:
  maxl=[]
  v=1
  for i in range(n+1):
    maxl.append(v)
    v=(v-b[i])*2
    if v<0:
      ret=-1
      break

v=0
vsum=0
if ret!=-1:
  for i in range(n,-1,-1):
    v=min(v+b[i],maxl[i])
    vsum+=v

if ret==-1:
  print(-1)
else:
  print(vsum)