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(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};
fn main() {
    input! {//
        n:usize,
        a:[i64;n+1],
    }
    if n == 0 {
        if a[0] == 1 {
            println!("{}", 1);
            return;
        }
    }

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

    let mut l = vec![0; n + 2];
    let mut r = vec![0; n + 2];
    l[n] = a[n];
    r[n] = a[n];
    for i in (0..n).rev() {
        l[i] = a[i] + (l[i + 1] + 1) / 2;
        r[i] = a[i] + r[i + 1];
    }
    if l[0] > 1 {
        println!("{}", -1);
        return;
    }
    let mut now = 1;
    let mut ans = 0;
    for i in 0..=n {
        ans += now;
        now = min((now - a[i]) * 2, r[i + 1]);
    }
    println!("{}", ans);
}