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: Time Limit Exceeded)

import java.io.*
import java.util.*
import kotlin.math.*

fun main(){
	val f = BufferedReader(InputStreamReader(System.`in`))

	val n = f.readLine().toInt()
	val array = f.readLine().split(" ").map{it.toLong()}

	var answer : Long = 1L

	val pq = PriorityQueue<Pair<Long,Long>>(compareBy({it.first}))
	if(array[0] > 1L){
		println(-1)
		return
	} else if(array[0] == 0L){
		pq.add(Pair(0L,0L))
		pq.add(Pair(0L,0L))
	}
	for(k in 1..n){
		if(array[k] == 0L) continue
		for(j in 1..array[k]){
			if(pq.isEmpty() || pq.peek().first >= k.toLong()){
				println(-1)
				return
			}

			val p = pq.poll()
			answer += k-p.first
			//println("${p.first} answer")
			if(p.second > p.first) pq.add(Pair(p.first+1,p.second))
			if(p.first+1 <= (k-1).toLong()) pq.add(Pair(p.first+1,(k-1).toLong()))
		}
	}

	println(answer)
}