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()))

c = 1
M = [] #（葉でない）子ノードの数の最大値
INF = 10**16
for a in A:
    c -= a
    M.append(c)
    c *= 2

    c = min(INF, c)
    
if c < 0:
    print(-1)
else:
    cnt = A[-1]
    num = A[-1]
    for a,m in zip(reversed(A[:-1]),reversed(M[:-1])):
        n = min(num, m)
        cnt += n+a
        num = n+a
    print(cnt)