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()))
b = [0]*(n+1)
if n == 0:  
    x = a[0]
    if x == 1:
        print(1)
    else:
        print(-1)
    exit()
nows = sum(a)
s = nows
for i in range(n+1):
    if i == 0:
        b[i] = 1 - a[i]
    else:
        nows -= a[i]
        b[i] = min(nows, 2*b[i-1]-a[i])
        if b[i] < 0:  
            print(-1)
            exit()

print(sum(b)+s)  
