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 kotlin.math.min

fun main() {
    val n = readLine()!!.toInt()
    val leaves = readLine()!!.split(" ").map { it.toLong() }
    if (n == 0) {
        if (leaves[0] == 1L) {
            println(1)
        } else {
            println(-1)
        }
        return
    }
    var curr = 0L
    for (j in n downTo 1) {
        curr += leaves[j]
        curr = (curr + 1L) / 2L
    }
    curr += leaves[0]
    if (curr != 1L) {
        println(-1)
        return
    }
    val dp = LongArray(n + 2)
    for (j in n downTo 0) {
        dp[j] = dp[j + 1] + leaves[j]
        if (j <= 60) {
            dp[j] = min(dp[j], 1L shl j)
        }
    }
    for (j in 1..n) {
        dp[j] = min(dp[j], 2L * (dp[j - 1] - leaves[j - 1]))
    }
    println(dp.sum())
}