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=input()
a=map(int,raw_input().split())

INF=10**18

pr=[0]*(n+2)

pr[0]=1

for i in xrange(n+1):
	pr[i+1]=min(INF,(pr[i]-a[i])*2)

if pr[n+1]<0:
	print "-1"
	exit(0)

s=0
ans=0

for i in xrange(n,-1,-1):
	s=min(s+a[i],pr[i])
	ans+=s

print ans