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)
  exit(0)
elif A[0] > 0:
  print(-1)
  exit(0)

X = [0] * (N+1)
X[-1] = A[-1]
for i in range(N,0,-1):
  X[i-1] = X[i] + A[i-1]
  
#print(X)

wk = 2
ans = 1
for i in range(1,N+1):
  
  #print(i,X[i])
  if 0 <= A[i] <= wk:
    ans += min(wk,X[i])
    wk -= A[i]
    wk *= 2
  else:
    print(-1)
    exit(0)
  
print(ans)