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;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using System.Text;


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        var n = long.Parse(Console.ReadLine());
        double[] a = Console.ReadLine().Split().Select(double.Parse).ToArray();

        double[] list = new double[n + 1];
        list[n] = a[n];
        for (long i = n; i > 0; i--)
        {
            list[i - 1] = list[i] + a[i - 1];
        }

        list[0] = 1;
        for (long i = 0; i < n; i++)
        {
            if ((list[i] - a[i])* 2 < list[i + 1])
            {
                list[i + 1] = (list[i] - a[i]) * 2;
            }
        }

        double ans = list.Sum();
        for (long i = 0; i < n + 1; i++)
        {
            if (list[i] - a[i] < 0)
            {
                ans = -1;
                break;
            }
        }

        // 出力
        Console.WriteLine(ans);
    }
}