Problem:
Score : 600 points

Problem Statement

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.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

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.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

def die; p(-1); exit; end

N,*L = $<.read.split.map(&:to_i)

B = [1-L[0]]+[0]*N
N.downto(1).inject(0){|s,i|
	next L[i] + B[i]=s
}
N.times{|d|
	b = B[d]*2 - L[d+1]
	die if b<0
	if b < B[d+1]
		B[d+1] = b
	else
		die if 0 < d.downto(1).inject(B[d]-B[d+1]-L[d+1]){|c,d|
			break c if c<=0
			B[d] -= c
			die if B[d]<=0
			next B[d-1]-B[d]-L[d]
		}
	end
}
die if B[0...N].any?(&:zero?)

p B.sum+L.sum