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()))

L = [0]*(n+2)
R = [0]*(n+2)
L[n] = A[n]
R[n] = A[n]

for i in range(n,-1,-1):
  L[i] = A[i] + (L[i+1]+1)//2
  R[i] = A[i] + R[i+1]
  
if L[0]>1:
  print(-1)
else:
  tmp = 1
  ans = 0
  for i in range(n+1):
    ans += tmp
    tmp -= A[i]
    tmp = min(tmp*2,R[i+1])    
  print(ans)    