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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;

namespace test
{
    class Program
    {
        static int[] ReadNumList()
        {
            var line = Console.ReadLine().Split(' ');
            var numList = new int[line.Length];
            for (var i = 0; i < line.Length; ++i)
            {
                numList[i] = int.Parse(line[i]);
            }
            return numList;
        }
        static void Main()
        {
            var n = int.Parse(Console.ReadLine());
            var a = ReadNumList();
            Console.WriteLine(Exec(a));
        }
        static long Exec(int[] a)
        {
            var maxAn = 1L;
            var points = new long[a.Length];
            // 各深さに存在しうる最大頂点数
            for (var d = 0; d < a.Length - 1; ++d)
            {
                points[d] = maxAn;
                maxAn -= a[d];
                if (maxAn < 0)
                {
                    return -1;
                }
                maxAn *= 2;
            }
            if (a[a.Length - 1] > maxAn)
            {
                return -1;
            }
            // 下から、条件を満たす中で最大の頂点数を設定
            points[a.Length - 1] = a[a.Length - 1];
            long total = points[a.Length - 1];
            for (var d = a.Length - 2; d >= 0; --d)
            {
                points[d] = Math.Min(points[d + 1] + a[d], points[d]);
                total += points[d];
            }
            return total;
        }
    }
}