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 std::io::{self, BufRead};
use std::cmp::min;

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let n: usize = lines.next().unwrap().unwrap().parse().unwrap();
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let mut paths: usize = 1;
  let mut vs: Vec<usize> = Vec::new();
  if n == 0 {
    println!("{}", if xs[0] == 1 { 1 } else { -1 });
    return;
  }
  for i in 0..n {
    if paths <= xs[i] {
      println!("-1");
      return;
    }
    vs.push(paths - xs[i]);
    paths = (paths - xs[i]) * 2;
  }
  if paths < xs[n] {
    println!("-1");
    return;
  }
  vs[n-1] = min(vs[n-1], xs[n]);
  for i in 1..n {
    vs[n-i-1] = min(vs[n-i-1], xs[n-i] + vs[n-1]);
  }
  let ans: usize = xs.iter().sum::<usize>() + vs.iter().sum::<usize>();
  println!("{}", ans);
}