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(a) for a in input().split()]

ans = 1
L = [0]*(N+2)
for i in range(N+1):
    L[i+1] = L[i]+A[i]

v = 1
for i in range(1, N+1):
    v = min(2*v, L[-1]-L[i])
    ans += v
    v -= A[i]
    if v < 0:
        ans = -1
        break
if A[0] > 0:
    ans = -1
    if N == 0 and A[0] == 1:
        ans = 1

print(ans)