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 proconio::input;

fn main() {
    input! {
        n: usize,
        a: [u64; n+1],
    }
    let mut inner_nodes = vec![0; n + 1];
    let nodes: Vec<_> = a
        .iter()
        .rev()
        .scan(0, |cum, &x| {
            *cum = *cum + x;
            Some(*cum)
        })
        .collect();
    let mut no_add = false;
    for (i, (&ai, &ni)) in a.iter().zip(nodes.iter().rev()).enumerate() {
        if !no_add && 2u64.pow(i as u32) > ni {
            no_add = true;
        }
        let same_depth = if no_add { ni } else { 2u64.pow(i as u32) };
        let prev_inner_nodes = if i > 0 { inner_nodes[i - 1] } else { 1 };
        if same_depth.min(prev_inner_nodes * 2) >= ai {
            inner_nodes[i] = same_depth.min(prev_inner_nodes * 2) - ai;
        } else if !(same_depth == ai && i == n) {
            println!("{}", -1);
            std::process::exit(0);
        }
    }
    if inner_nodes[0] == 1 {
        println!(
            "{}",
            inner_nodes
                .iter()
                .zip(a.iter())
                .fold(0, |sum, (n, a)| sum + n + a)
        );
    } else {
        println!("{}", -1);
    }
}