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_imports)]
use itertools::Itertools;
use proconio::{input, marker::*};
use std::cmp::*;
use std::collections::*;
use superslice::*;

fn main() {
    input! {
        n: usize,
        a: [i64;n+1],
    }
    let ans = dfs(a[n], n, &a);
    println!("{}", ans);
}

fn dfs(v: i64, depth: usize, a: &Vec<i64>) -> i64 {
    if v > 2i64.pow(depth as u32) {
        return -1;
    }

    if depth == 0 {
        return 1;
    }

    let mut res = -1;

    for i in (v + 1) / 2..=v {
        let next = dfs(i + a[depth - 1], depth - 1, a);
        res = max(res, next);
    }

    res = if res == -1 { res } else { res + v };
    res
}