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: Accepted)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.{Map, Stack, Queue, PriorityQueue, Set}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val N = in.next().toInt
  val A = new Array[Long](N+1)
  for (i <- 0 to N) {
    A(i) = in.next().toLong
  }

  //
  var flag = true
  val dpMax = new Array[Long](N+1)
  var cnt = 1L
  breakable {
    for (i <- 0 to N) {
      cnt -= A(i)
      if (cnt < 0) {
        flag = false
        break
      }

      dpMax(i) = cnt
      cnt *= 2
      cnt = cnt min (Long.MaxValue / 100)
    }
  }


  if (!flag)
    println(-1)
  else if (N == 0) {
    if (A(0) == 1)
      println(1)
    else
      println(-1)
  }
  else  {
    var ans = 0L
    var nodes = 0L

    breakable {
      for (i <- (0 to N).reverse) {
        val now = A(i)

        if ((nodes+1) / 2 > dpMax(i)) {
          flag = false
          break
        }
        else {
          nodes = (nodes min dpMax(i)) + now
          ans += nodes
        }
      }
    }

    if (!flag)
      println(-1)
    else
      println(ans)
  }
}

object Sub {
  def print2DimArray[A](mat: Array[Array[A]]): Unit = {
    mat.foreach(lin => println(lin.toList))
  }
}

class InputReader(stream: InputStream) {
  private[this] val reader = new BufferedReader(new InputStreamReader(stream))
  private[this] var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}