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)
B[N]=A[N]
for i in range(N-1,-1,-1):
    B[i]=B[i+1]+A[i]

C=[0]*(N+1)
C[0]=1
for i in range(1,N+1):
    C[i]=2*(C[i-1]-A[i-1])

check=[min(B[i],C[i]) for i in range(N+1)]
for i in range(N+1):
    if check[i]<A[i]:
        print(-1)
        exit()
print(sum(check))