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())
v = list(map(int, input().split()))
u = [1]*(n+1)
a = 0
b = False
tmp = 0
tt = 0
for i in range(n+1):
  if u[i] < v[i]:
    b = True
    break
  if i < n:
    u[i+1] = (u[i] - v[i])*2
for i in range(n, -1, -1):
  tt += v[i]
  tmp = min(tt, u[i])
  a += tmp
  tt = tmp
if n == 0:
  a = v[0]
if b:
  a = -1
print(a)