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)

n = gets.to_i
as = gets.split.map &:to_i

if n == 0 then
	if as[0] == 1 then
		puts 1
		exit
	else
		puts -1
		exit
	end
end

start = 1

fl = [1 - as[0]]

(1..n).each{ |x|
	start = start * 2 - as[x]
	if start < 0 then
		puts -1
		exit
	end
	
	fl << start
}

if as[-1] == 0 || as[-1] > fl[-1] then
	puts -1
	exit
end

sum = as[-1]

next_ = as[-1]
(as.size - 2).downto(0){ |i|
	if next_ < fl[i] then
		sum += next_ + as[i]
		next_ += as[i]
	else
		sum += fl[i] + as[i]
		next_ = fl[i] + as[i]
	end
}

puts sum