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(raw_input())
ais = map(int, raw_input().split(' '))

def f(ais,n):
	q = 1 # number of non leaves
	r = 0
	h = {}
	for depth,nl in enumerate(ais):
		if nl > q: return -1
		elif nl == q:
			if depth < n: return -1
			q -= nl
		elif nl < q:
			q -= nl
		#r += (q + nl)
		h[depth] = q
		if depth < n: q *= 2
	
	internal = 0
	for j in range(len(ais)-1, 0, -1):
		r += ais[j] + internal
		internal = min(h[j - 1], ais[j] + internal) 
	return r + 1


print f(ais,n)

