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 Rust (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use text_io::*;

fn main() {
    let n: usize = read!();
    let mut leafs: Vec<isize> = Vec::with_capacity(n);
    let mut max = 1;
    for i in 0..=n {
        let a = read!();
        leafs.push(a);
        if i < 27 && leafs[i] > max {
            println!("-1");
            std::process::exit(0);
        } else {
            max *= 2;
        }
    }

    if leafs[0] != 0 && n != 0 {
        println!("-1");
        std::process::exit(0);
    }

    let mut up: Vec<(isize, isize)> = vec![];
    up.push((1, 1));
    for i in 1..=n {
        let min = up[i - 1].0 - leafs[i];
        let max = (up[i - 1].1 * 2 - leafs[i]).min(2usize.pow(i as u32) as isize - leafs[i]);
        up.push((min, max));
    }

    let mut down: Vec<(isize, isize)> = vec![];
    down.push((0, 0));
    for i in 0..n {
        let m = n - i;
        let min = (leafs[m] + down[i].0 + 1) / 2;
        let max = (leafs[m] + down[i].1).min(2usize.pow((m - 1) as u32) as isize - leafs[m - 1]);
        down.push((min, max));
    }
    down.reverse();

    let mut ans = vec![];
    for i in 0..=n {
        let min_up = up[i].0;
        let min_down = down[i].0;
        let max_up = up[i].1;
        let max_down = down[i].1;
        if max_up < min_down || min_up > max_down {
            println!("-1");
            std::process::exit(0);
        } else {
            ans.push(max_up.min(max_down));
        }
    }
    // println!("{:?}\n{:?}\n{:?}", leafs, up, down);

    let ans: isize = leafs.iter().sum::<isize>() + ans.iter().sum::<isize>();
    println!("{}", ans);
}