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_parens)]
#![allow(unused_imports)]
#![allow(non_upper_case_globals)]
#![allow(non_snake_case)]
#![allow(unused_mut)]
#![allow(unused_variables)]
#![allow(dead_code)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};

#[allow(unused_macros)]
#[cfg(debug_assertions)]
macro_rules! mydbg {
    //($arg:expr) => (dbg!($arg))
    //($arg:expr) => (println!("{:?}",$arg));
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
      }
}
#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::*;
use std::collections::*;
use std::ops::{Add, Div, Mul, Sub};

#[allow(dead_code)]
static INF_I64: i64 = 92233720368547758;
#[allow(dead_code)]
static INF_I32: i32 = 21474836;
#[allow(dead_code)]
static INF_USIZE: usize = 18446744073709551;
#[allow(dead_code)]
static M_O_D: usize = 1000000007;
#[allow(dead_code)]
static PAI: f64 = 3.1415926535897932;

trait IteratorExt: Iterator {
    fn toVec(self) -> Vec<Self::Item>;
}

impl<T: Iterator> IteratorExt for T {
    fn toVec(self) -> Vec<Self::Item> {
        self.collect()
    }
}

trait CharExt {
    fn toNum(&self) -> usize;
    fn toAlphabetIndex(&self) -> usize;
    fn toNumIndex(&self) -> usize;
}
impl CharExt for char {
    fn toNum(&self) -> usize {
        return *self as usize;
    }
    fn toAlphabetIndex(&self) -> usize {
        return self.toNum() - 'a' as usize;
    }
    fn toNumIndex(&self) -> usize {
        return self.toNum() - '0' as usize;
    }
}

trait VectorExt {
    fn joinToString(&self, s: &str) -> String;
}
impl<T: ToString> VectorExt for Vec<T> {
    fn joinToString(&self, s: &str) -> String {
        return self
            .iter()
            .map(|x| x.to_string())
            .collect::<Vec<_>>()
            .join(s);
    }
}

trait StringExt {
    fn get_reverse(&self) -> String;
}
impl StringExt for String {
    fn get_reverse(&self) -> String {
        self.chars().rev().collect::<String>()
    }
}

trait UsizeExt {
    fn pow(&self, n: usize) -> usize;
}
impl UsizeExt for usize {
    fn pow(&self, n: usize) -> usize {
        return ((*self as u64).pow(n as u32)) as usize;
    }
}

fn main() {
    input! {
       mut N: usize,
       a: [usize;N+1],
    }
    N += 1;

    let mut tmp = vec![(0, 0); N];
    tmp[0] = (1, 1);
    tmp[N - 1] = (a[N - 1], a[N - 1]);
    for i in (1..N - 1).rev() {
        let b = a[i];
        let (c, d) = tmp[i + 1];
        let e = (c + 1) / 2 + b;
        let f = d + b;
        tmp[i] = (e, f);
    }
    mydbg!(tmp);
    let mut ans = vec![0; N];
    ans[0] = 1;
    for i in 1..N {
        let b = ans[i - 1];
        let (c, d) = tmp[i];
        if b <= a[i - 1] {
            echo!(-1);
            return;
        }
        if b * 2 < c {
            echo!(-1);
            return;
        }
        let e = min((b - a[i - 1]) * 2, d);
        let f = max(c, e);
        ans[i] = f;
    }

    mydbg!(ans);
    echo!(ans.iter().fold(0, |a, x| a + x));
}