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: Accepted)

#![allow(unused_imports)]
#![allow(non_snake_case, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

// https://atcoder.jp/contests/hokudai-hitachi2019-1/submissions/10518254

macro_rules! eprint {
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = write!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! eprintln {
	() => { eprintln!(""); };
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = writeln!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! dbg {
	($v:expr) => {{
		let val = $v;
		eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
		val
	}}
}

macro_rules! mat {
	($($e:expr),*) => { Vec::from(vec![$($e),*]) };
	($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
	($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
	($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
	($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
		$a$([$i])*.$f($($t),*)
	};
	($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
		let t = $e;
		ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
	} };
}

pub fn readln() -> String {
	let mut line = String::new();
	::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
	line
}

macro_rules! read {
	($($t:tt),*; $n:expr) => {{
		let stdin = ::std::io::stdin();
		let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
			let line = line.unwrap();
			let mut it = line.split_whitespace();
			_read!(it; $($t),*)
		}).collect::<Vec<_>>();
		ret
	}};
	($($t:tt),*) => {{
		let line = readln();
		let mut it = line.split_whitespace();
		_read!(it; $($t),*)
	}};
}

macro_rules! _read {
	($it:ident; [char]) => {
		_read!($it; String).chars().collect::<Vec<_>>()
	};
	($it:ident; [u8]) => {
		Vec::from(_read!($it; String).into_bytes())
	};
	($it:ident; usize1) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
	};
	($it:ident; [usize1]) => {
		$it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
	};
	($it:ident; [$t:ty]) => {
		$it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
	};
	($it:ident; $t:ty) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
	};
	($it:ident; $($t:tt),+) => {
		($(_read!($it; $t)),*)
	};
}


pub fn main() {
	let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: i64 = 1_000_000_007;
const INF: i64 = std::i64::MAX/3;

fn solve() {
	let n = read!(usize);
	let mut a = read!([i64]);
	let mut cum = vec![0;n+2];
	for i in 0..n+1 {
		cum[i+1] = cum[i]+a[i];
	}
	let mut prev = 1;
	let mut ans = 0;
	for i in 0..n+1 {
		let mut ma = prev*2-a[i];
		if i==0 {
			ma = 1-a[i];
		}
		ma = min(ma, cum[n+1]-cum[i+1]);
		if ma < 0 {
			println!("-1");
			return;
		}
		ans += ma+a[i];
		prev = ma;
	}
	println!("{}",ans);
}

fn run() {
    solve();
}