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)

import java.util.*

fun main(args: Array<String>) {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toInt()
        val A = Array(N + 1) { sc.next().toInt() }
        if (A[0] != 0) {
            pw.println("-1")
            return
        }
        var q = mutableSetOf<Pair<Long, Int>>()
        q.add(Pair(A[N].toLong(), A[N]))
        for (i in N - 1 downTo 0) {
            //dprint(q)
            val q2 = mutableSetOf<Pair<Long, Int>>()
            q.forEach {
                for (j in (it.second + 1) / 2..it.second) {
                    q2.add(Pair(it.first + j + A[i], j + A[i]))
                }
            }
            q = q2
        }
        //dprint(q)

        val ans = q.filter { it.second == 1 }.maxBy { it.first }
        pw.println(ans?.first ?: "-1")
    }
}