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(args: Array<String>) {
    fun divWithCeil(a: Long, b: Long): Long {
        return (a + b - 1) / b
    }

    val n = readLine()!!.toInt()
    val aList = readLine()!!.split(' ').map(String::toLong)
    if (n == 0) {
        if (aList[0] == 1L) {
            println(1)
        } else {
            println(-1)
        }
        return
    }

    val nodeNumArray = LongArray(n + 2) { 0L }
    nodeNumArray[0] = 1
    val maxNodeNumArray = LongArray(n + 2) { 1 }
    (1..n).forEach {
        maxNodeNumArray[it] = if (it > 30) Long.MAX_VALUE / 2 else (maxNodeNumArray[it - 1] - aList[it - 1]) * 2
    }

    for (i in n downTo 1) {
        val a = aList[i]
        val beforeA = aList[i - 1]
        val nodeMax = maxNodeNumArray[i]
        val minUsedNode = divWithCeil(nodeNumArray[i + 1], 2L)
        val maxUsedNode = nodeNumArray[i + 1]
        if (a + minUsedNode > nodeMax) {
            println(-1)
            return
        } else if (a + maxUsedNode > nodeMax) {
            nodeNumArray[i] = nodeMax
        } else {
            nodeNumArray[i] += a
            nodeNumArray[i] += nodeNumArray[i + 1]
        }
    }
    println(nodeNumArray.sum())
}