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 = [int(k) for k in input().split()]


M = [1]
for a in A[:-1]:
    M.append((M[-1] - a) * 2)


dsum = 0
d = 0
ok = True
for i in range(len(A) - 1, -1, -1):
    min_d = ((d + 1) // 2 if d % 2 else d // 2) + A[i]
    max_d = d + A[i]
    if min_d > M[i]:
        ok = False
        break
    d = min(M[i], max_d)
    dsum += d

print(dsum if ok else -1)
