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
import queue
n = int(input())
a = list(map(int,input().split()))
xx = [0]*(n+1)
x = 1
c = 0
for i in range(n+1):
  if x >= a[i]:
    c += a[i]
    xx[i] = x-a[i]
    x = 2*(x-a[i])
  else:
    print(-1)
    exit()
y = a[n]
i = n-1
while y > 0:
  if xx[i] >= y:
    c += y
  else:
    c += xx[i]
    y = xx[i]
  y += a[i]
  if i > 0:
    i -= 1
  else:
    break
print(c)