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: Time Limit Exceeded)

def main
    n = gets.to_i
    cs = gets.chomp.split(" ").map(&:to_i)
    p_max = [1]
    p_min = [1000000000] * n + [0]
    ps = []
    sums = []

    (1..n).each do |i|
        p_max.push (p_max[i-1] * 2 - cs[i])
        if p_max[i] < 0
            puts -1
            return
        end
    end
    (0..n-1).reverse_each do |i|
        p_min[i] = p_min[i+1] + cs[i+1]
        break if p_min[i] > p_max[i]
    end
    (0..n).each do |i|
        ps.push [p_max[i], p_min[i]].min
        sums.push (ps[i] + cs[i])
        if sums[i] > 2**i
            puts -1
            return
        end
    end
    puts sums.sum
end

main