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
def f(N, A):
    B = [0]*(N+1)
    b = 0
    for i in range(N, -1, -1):
        b += A[i]
        B[i] = b


    x = 1
    s = 0

    for i in range(N+1):
        x = min(x, B[i])
        s += x
        
        x = (x - A[i])*2
        if x < 0:
            return -1
        
    if x != 0:
        return -1

    return s


N = int(input())
A = list(map(int, input().split()))
print(f(N, A))
