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)

use proconio::{fastout, input};
use std::cmp::min;

#[fastout]
fn main() {
    input! {in_n: usize, in_a: [u64; in_n + 1]};
    let mut leaves = vec![0 as u64; in_n + 1];
    let depth = in_n + 1;

    if in_a[0] != 0 {
        println!("-1");
        return;
    }

    for i in (2..depth).rev() {
        leaves[i - 1] = leaves[i] + in_a[i];
    }

    let mut node = 1;
    let mut total = 1;

    for i in 1..depth {
        let n = node * 2;

        if in_a[i] > n {
            println!("-1");
            return;
        }

        let r = min(n - in_a[i], leaves[i]);
        total += in_a[i] + r;
        node = r;
    }

    println!("{}", total);
}