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(non_snake_case)]

use std::cmp::min;
use std::io::{stdin, Read};
use std::str::FromStr;
fn read_option<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok()
}
fn read<T: FromStr>() -> T {
    let opt = read_option();
    opt.expect("failed to parse token")
}

fn main() {
    let N: usize = read();
    let vec: Vec<_> = (0..=N).map(|_| read::<u64>()).collect();
    if N == 0 && vec[0] == 1 {
        println!("1");
        return;
    }
    if vec[0] != 0 {
        println!("-1");
        return;
    }
    let mut before = 1;
    for &v in &vec {
        if v > before {
            println!("-1");
            return;
        }
        before -= v;
        before *= 2;
    }
    let mut sum: u64 = vec.iter().sum();
    let mut ans: u64 = 1;
    let mut now_ver: u64 = 1;
    for &v in vec.iter().skip(1) {
        now_ver = min(now_ver.saturating_mul(2), sum);
        ans += now_ver;
        now_ver -= v;
        sum -= v;
    }
    println!("{}", ans)
}