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.Linq;
using System.Security.Cryptography.X509Certificates;
using System.Text;
using System;
using System.Collections;
using System.Collections.Generic;
using System.Globalization;
using System.IO;

namespace AtCoder
{
    class Program
    {

        const int BiggestMod = 1000000007;
        public static void Main()
        {

            var nums = int.Parse(Console.ReadLine()) + 1;
            var list = new KeyValuePair<long, long>[nums]; // Key = 枝, Value = 葉
            var args = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();

            for (int i = args.Length - 1; args[i] == 0; i--)
                Array.Resize(ref args, args.Length - 1);

            if (args[0] == 0)
            {
                list[0] = new KeyValuePair<long, long>(1, 0);
                if (args.Length == 1)
                {

                    Console.WriteLine(-1);
                    return;

                }
            }
            else if (args[0] == 1) list[0] = new KeyValuePair<long, long>(0, 1);
            else
            {

                Console.WriteLine(-1);
                return;

            }

            for (int i = 1; i < nums; i++)
            {

                if ((list[i - 1].Key << 1) < args[i])
                {

                    Console.WriteLine(-1);
                    return;

                }
                
                else
                {

                    var leaf = args[i];
                    var branch = (list[i - 1].Key << 1) - args[i];
                    if (branch > 2000000000) branch = 2000000000;

                    list[i] = new KeyValuePair<long, long>(branch, leaf);
                }

            }

            list[nums - 1] = new KeyValuePair<long, long>(0, args[nums - 1]);
            
            for (int i = nums - 2; 
                i >= 0; i--)
            {

                if (list[i].Key > (list[i + 1].Key + list[i + 1].Value))
                    list[i] = new KeyValuePair<long, long>(list[i + 1].Value + list[i + 1].Key, list[i].Value);

            }

            Console.WriteLine(list.Sum(x => x.Key + x.Value));


        }

    }

}