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.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
using Library;
using static Library.Input;

namespace AtCoderTemplate
{
    class ProgramC
    {
        static void Main()
        {
            int n = NextInt;
            var a = LongList;
            var array = Enumerable.Repeat(long.MaxValue >> 2, n + 1).ToArray();
            array[0] = 1;

            for (int i = 0; i < n && array[i] <= long.MaxValue >> 4; i++)
            {
                array[i + 1] = (array[i] - a[i]) * 2;
            }

            if (array.Any(x => x <= 0) || (n == 0 && a[0] != 1))
            {
                System.Console.WriteLine(-1);
                return;
            }

            array[n] = Math.Min(a[n], array[n]);
            for (int i = n - 1; i >= 0 ; i--)
            {
                array[i] = Math.Min(array[i], array[i + 1] + a[i]);
            }

            // System.Console.WriteLine(string.Join(" ", array));
            System.Console.WriteLine(array.Sum());
        }
    }
}

namespace Library
{
    class Input
    {
        static IEnumerator<string> enumerator = new string[] { }.AsEnumerable().GetEnumerator();

        public static string Line => Console.ReadLine();

        public static string[] StrArr => Line.Split(' ');

        public static int NextInt => int.Parse(NextWord());

        public static long NextLong => long.Parse(NextWord());

        public static List<int> IntList => StrArr.Select(int.Parse).ToList();

        public static List<long> LongList => StrArr.Select(long.Parse).ToList();

        public static IEnumerable<long[]> TakeLine(int N)
        {
            return Enumerable.Repeat(0, N).Select(_ => Console.ReadLine().Split(' ').Select(long.Parse).ToArray());
        }

        public static string NextWord()
        {
            while (!enumerator.MoveNext())
            {
                enumerator = StrArr.AsEnumerable().GetEnumerator();
            }
            return enumerator.Current;
        }
    }
}