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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main() {
  val n = readLine()!!.toInt()
  val a = readLine()!!.split(' ').map { it.toInt() }.toIntArray()
  var min = LongArray(n + 1)
  val max = LongArray(n + 1)
  min[n] = a.last().toLong()
  max[n] = a.last().toLong()

  fun limit(level: Int, vertices: Long): Long = if (level > 63) vertices else kotlin.math.min(1L shl level, vertices)

  for (i in n - 1 downTo 0) {
    val mn = (min[i + 1] + 1) / 2 + a[i]
    val mx = limit(i,max[i + 1] + a[i])

    if (mn > mx) {
      println(-1)
      return
    }

    min[i] = mn
    max[i] = mx
  }

  for (i in 1..n) {
    if ((max[i] + 1) / 2 > max[i - 1] - a[i - 1]) {
      max[i] = kotlin.math.min(max[i], (max[i - 1] - a[i - 1]) * 2)
    }
    if (max[i] < min[i]) {
      println(-1)
      return
    }
  }
  check(max.find { it < 0 } == null)

  if (max[0] != 1L) {
    println(-1)
    return
  }

  println(max.sum())
}