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: Accepted)

import java.lang.Long.*

inline fun str() = readLine()!!
inline fun int() = str().toInt()
inline fun long() = str().toLong()

inline fun read(d:String = " ") = str().split(d)
inline fun <T> list(d:String = " ",f:String.()->T) = read(d).map {f(it)}
inline fun list() = list {toLong()}.toLongArray()

inline fun <T> T.print() = this.also(::println)

fun main() {
	abc000X()
}

fun abc000X() {
	val n = int()
	val leaf = list()
	val notleaf = LongArray(n+1) {0L}
	val leafsum = LongArray(n+1) {0L}

	fun lim(i:Int) = if(i>=60) Long.MAX_VALUE else 1L shl i

	for(i in n downTo 0) {
		if(leaf[i]+notleaf[i]>lim(i)) {
			println(-1)
			return
		}

		if(i>0) {
			notleaf[i-1] += (leaf[i]+notleaf[i]+1)/2
			leafsum[i-1] += leafsum[i]+leaf[i]
		}
	}

	for(i in 1..n)
		notleaf[i] = min(notleaf[i-1]*2-leaf[i],leafsum[i])

	//	System.err.println(leaf.joinToString())
	//	System.err.println(notleaf.joinToString())
	//	System.err.println(sum.joinToString())
	println(leaf.sum()+notleaf.sum())
}