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)

use std::cmp::min;
use std::io::stdin;

fn main() {
    let mut is = String::new();
    stdin().read_line(&mut is).ok();
    let n: usize = is.trim().parse().unwrap();
    is.clear();
    stdin().read_line(&mut is).ok();
    let a: Vec<usize> = is.split_whitespace().map(|e| e.parse().unwrap()).collect();
    is.clear();
    let mut min_nodes = vec![0; n + 2];
    let mut max_nodes = vec![0; n + 2];
    let mut flg;
    for i in (0..=n).rev() {
        min_nodes[i] = (min_nodes[i + 1] + 1) / 2 + a[i];
        max_nodes[i] = max_nodes[i + 1] + a[i];
    }
    let mut ans = 0;
    let mut now = 0;
    flg = max_nodes[0..max_nodes.len() - 1].iter().all(|&e| e > 0);
    for i in 0..=n {
        let next_min = min_nodes[i];
        let next_max = max_nodes[i];
        now = min(next_max, (now - if i > 0 { a[i - 1] } else { 0 }) * 2);
        ans += now;
        // now = min((now - a[i]) * 2, next_max);
        if now < next_min || flg == false {
            flg = false;
            break;
        }
    }

    if flg {
        println!("{}", ans);
    } else {
        println!("-1");
    }
}