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())
A = list(map(int, input().split()))

if N == 0 and A[0] != 1:
  print(-1)
elif N != 0 and A[0] != 0:
  print(-1)
else:
  ans = 1
  nodes = 1
  preF = 0
  sucF = sum(A)
  
  for d in range(1, N+1):
    fol = A[d]
    nodes = min((nodes-preF)*2, sucF)
    
    if nodes < fol:
      print(-1)
      exit()
    
    ans += nodes
    preF = fol
    sucF -= fol
  print(ans)