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() {
    val n = readLine()!!.toInt()
    val an = readLine()!!.split(" ").map(String::toInt)
    val dp = IntArray(n + 1)
    dp[an.lastIndex] = an.last()
    for (i in an.lastIndex - 1 downTo 0) {
        dp[i] = Math.max(dp[i + 1], an[i])
    }
    dp[an.lastIndex] = 0

    // 根がおかしい場合
    if (0 < n) {
        if (an[0] == 1) {
            println(-1)
            return
        }
    }

    // 根はループで処理しない
    var count = 1
    var previousValue = 1

    for (i in 1 until an.lastIndex) {
        if (previousValue * 2 < an[i]) {
            println(-1)
            return
        }
        val maxCurrentValue = previousValue * 2 - an[i]
        val currentValue = Math.min(maxCurrentValue, dp[i + 1] + an[i + 1])
//        println("$currentValue ${an[i]}")
        count += (currentValue + an[i])
        previousValue = currentValue
    }

    count += an.last()
    println(count)
}