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 OCaml (4.10.0) (judge verdict: Wrong Answer)

let () = Scanf.scanf "%d\n" @@ fun n ->
  let as_ = Array.init (n + 1) @@ fun _ -> Scanf.scanf "%d " @@ fun a -> a in
  let maximum = Array.make (n + 1) 0 in
  maximum.(0) <- max 0 (1 - as_.(0));
  for i = 0 to n - 1 do
    maximum.(i + 1) <- max 0 @@ 2 * maximum.(i) - as_.(i + 1)
  done;
  try
    let nodes = ref 0 in
    let leafs = ref as_.(n) in
    for i = n - 1 downto 0 do
      let leafs' = min !leafs maximum.(i) in
      if !leafs < leafs' || 2 * leafs' < !leafs then raise Not_found;
      nodes := !nodes + leafs';
      leafs := leafs'
    done;
    Printf.printf "%d\n" @@ !nodes + Array.fold_left ( + ) 0 as_
  with Not_found -> print_endline "-1"