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::str::FromStr;
use std::fmt::Debug;
use std::cmp;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()
}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main(){
    let n = read_as_vec::<usize>()[0];
    let mut a = read_as_vec::<usize>();

    let mut limits: Vec<usize> = vec![0;n+1];
    limits[0] = 1;
    for i in 1..limits.len(){
        if limits[i-1] < a[i-1]{
            println!("-1");
            return;
        }
        limits[i] = (limits[i-1] - a[i-1])*2;
    }

    //println!("imites = {:?}", limits);

    for i in (1..n+1).rev(){
        if a[i-1] + (a[i]+1)/2 > limits[i-1]{
            println!("-1");
            return;
        }

        if a[i-1] + a[i] <= limits[i-1]{
            a[i-1] += a[i];
        }else{
            a[i-1] = limits[i-1];
        }
        //println!("after i = {}, a= {:?}", i, a);
    }
    println!("{}", a.iter().sum::<usize>());
}