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)

use proconio::input;
use std::cmp::min;

fn main() {
    input! {
        n: u64,
        a: [u64; n+1]
    };
    if n == 0 {
        if a[0] == 1 {
            println!("1");
        } else {
            println!("-1");
        }
        return ();
    }

    let mut e: Vec<u64> = vec![0; (n+1) as usize];
    let mut s: u64 = a.iter().sum();
    e[0] = 1;
    let mut x: u64;

    for i in (1..(n+1)).map(|e| e as usize) {
        e[i] = min((e[i-1] - a[i-1]) * 2, s);
        if e[i] < a[i] {
            println!("-1");
            return ();
        }
        s = s - a[i];
    }
    println!("{}", e.iter().sum::<u64>());
}