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()))
isOK = True
B = [0]*(N+1)
B[0] = 1-A[0]
M = 1
i = 1
while i <= N:
  B[i] = B[i-1]*2-A[i]
  i += 1
if B[-1] < 0 or N > 0 and A[-1] == 0:
  isOK = False
B[-1] = A[-1]
i = N-1
while i >= 0:
  if B[i] < 1 or i > 0 and B[i] > B[i-1]*2:
    isOK = False
    break
  else:
    B[i] = min(B[i], B[i+1])+A[i]
  i -= 1
print(-1 if not isOK else sum(B))
