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()))

# 各階層で可能な点のmin、max
v=[[0,0] for i in range(n+1)]
v[n]=[a[-1],a[-1]]
for i in range(n-1,-1,-1):
    v[i][0]=-(-v[i+1][0]//2)+a[i]
    v[i][1]=v[i+1][1]+a[i]


if v[0][0]>1:
    print(-1)
    exit()

ans=1
# w＝各階層で葉をもつ頂点の個数
w=1
# 各階層でできるだけ多く点をとる。
for i in range(1,n+1):
    w=min(w*2,v[i][1])
    ans+=w
    # 葉っぱは次の階層の根にならない
    w-=a[i]
print(ans)