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>) {
    val n = readLine()!!.toInt()
    val a = readLine()!!.split(" ")

    var preh = 1L
    var pret = a[n].toLong()
    var max = LongArray(n+1)
    var min = LongArray(n+1)
    max[0] = preh
    min[n] = pret
    for (i in 1 .. n) {
        val ah = a[i-1].toLong()
        val at = a[n-i].toLong()
        preh = 2*(preh - ah)
        max[i] = preh
        min[n-i] = pret + at
        pret += at
    }
    if (preh <= 0L) {
        println(-1)
    } else {
        var ans = 0L
        for (i in 0..n) {
            ans += Math.min(max[i], min[i])
        }
        println(ans)
    }
}