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: Accepted)

n = gets.strip.to_i
a = gets.strip.split.map(&:to_i)

vmin = Array.new(n+2,0)
vmax = Array.new(n+2,0)
n.downto(0) do |i|
    vmin[i] = a[i]+(vmin[i+1]+1)/2
    vmax[i] = a[i]+vmax[i+1]
end
if vmin[0] > 1
    p -1
    exit
end
vmax[0] = 1
1.upto(n) do |i|
    vmax[i] = [vmax[i], (vmax[i-1]-a[i-1])*2].min
end
ans = 0
(n+1).times {|i| ans+= vmax[i]}
p ans