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: Time Limit Exceeded)

#[allow(unused)]
use proconio::marker::*;
use proconio::{fastout, input};
#[allow(unused)]
use std::cmp::{max, min};
#[allow(unused)]
use std::collections::{HashMap, HashSet, VecDeque};

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [i64; n+1],
    }
    if a[0] != 0 {
        if n == 0 && a[0] == 1 {
            println!("{}", 1);
        } else {
            println!("{}", -1);
        }
        return;
    }

    let mut rec = vec![vec![0, 1, 0]; n + 1];
    rec[n][0] = 1;
    rec[n][1] = 0;

    let mut j = 0;
    for (i, &a) in a.iter().enumerate() {
        // println!("{}: {:?}", j, rec);
        let mut a = a;
        if i == 0 || a == 0 {
            continue;
        }
        if i < n {
            rec[i][0] = a;
        } else {
            if a == 1 {
                break;
            }
            rec[n][0] = a;
            a -= 1;
        }

        while j < n + 1 && rec[j][1] == 0 {
            j += 1;
        }
        if j >= n + 1 {
            println!("{}", -1);
            return;
        }

        let mut b = a;
        let mut cnt = 0;
        let mut k = j;
        while k < i {
            // SLOW...
            rec[k][1] += cnt;
            if rec[k][1] <= b && b > 0 {
                rec[k][2] += rec[k][1];
                cnt += rec[k][1];
                b -= rec[k][1];
                rec[k][1] = 0;
                j += 1;
            } else {
                rec[k][1] -= b;
                rec[k][2] += b;
                cnt += b;
                b = 0;
            }
            k += 1;
        }
        if b != 0 {
            println!("{}", -1);
            return;
        }
    }
    // println!("{}: {:?}", j, rec);
    println!(
        "{}",
        rec.iter().fold(0 as i64, |s, r| s + r.iter().sum::<i64>())
    );
}