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 Scala (2.13.1) (judge verdict: Wrong Answer)

import java.io.{BufferedOutputStream, BufferedReader, InputStreamReader, PrintWriter}

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val n = in.readLine().toInt
    val as = in.readLine().split("\\s+").map(_.toInt)

    val maxes = Array.ofDim[Long](as.length)
    var sum = 0L
    for (i <- maxes.length - 1 to 1 by -1) {
      sum += as(i)
      maxes(i - 1) = sum
    }

    var count = 0L
    var node = 1L
    for (i <- as.indices) {
      if (node < as(i)) {
        out.println(-1)
        out.flush()
        return
      }
      count += Math.min(node, maxes(i) + as(i))
      node = (node - as(i)) * 2
    }

    out.println(count)
    out.flush()
  }
}