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;

fn main() {
    if let Some(ans) = solve() {
        println!("{}", ans)
    } else {
        println!("-1")
    }
}

fn solve() -> Option<u64> {
    input!(n: usize, a: [u64; n + 1]);
    if n == 0 {
        return if a[0] <= 1 { Some(a[0]) } else { None };
    }
    let mut s = a.clone();
    for i in (1..=n).rev() {
        s[i - 1] += s[i];
    }
    let mut b = vec![0; n];
    b[0] = 1u64.checked_sub(a[0])?;
    for i in 1..n {
        let bmax = (b[i - 1] * 2).checked_sub(a[i])?.min(s[i + 1]);
        let bmin = b[i - 1].saturating_sub(a[i]);
        if bmin > bmax {
            return None;
        }
        b[i] = bmax;
    }
    if n > 0 && a[n] <= b[n - 1] * 2 {
        Some(a.iter().sum::<u64>() + b.iter().sum::<u64>())
    } else {
        None
    }
}