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)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout()]
fn main() {
    input! {
        n: i64,
        aa: [i64; n as usize + 1]
    }

    let mut memo = HashSet::new();
    let ans = dfs(&aa, n, 0, 0, &mut memo);
    println!("{}", ans);


}

fn dfs(aa: &[i64], d: i64, parent: i64, sum: i64, memo: &mut HashSet<(i64, i64)>) -> i64 {
    if memo.contains(&(d, parent)) {
        return -1;
    }
    let num = parent + aa[d as usize];
    if num > 1<<d {
        return -1;
    }
    if d == 0 {
        return num + sum;
    }
    let mut low = (num+1) / 2;
    let mut up = min(num ,1 << (d-1)) + 1;
    let mut res = -1;
    let mut modified = false;
    while up - low > 1 {
        modified = false;
        let mid = (low + up) / 2;
        let r = dfs(aa, d-1, mid, num+sum, memo);
        res = max(r, res);
        if r == -1 {
            memo.insert((d-1, mid));
            up = mid;
        } else {
            low = mid;
            modified = true;
        }
    }
    if !modified {
        res = dfs(aa, d-1, low, num+sum, memo);
    }
    res

}