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: Runtime Error)

using System;
using System.Linq;

namespace AtCoder.Question
{
    public class QuestionC
    {
        public static void Main(string[] args)
        {
            var n = ReadValue<int>();
            var array = ReadValues<int>();
            long total = 0;
            long numberOfBranch = 1;
            long numberOfLeaves = array.Sum();
            for (int d = 0; d <= n; d++)
            {
                var leaves = array[d];
                var branch = d == 0 ? 1 : Math.Min(2 * numberOfBranch, numberOfLeaves);

                total += branch;

                numberOfBranch = branch - leaves;
                numberOfLeaves -= leaves;
                if (numberOfBranch < 0)
                {
                    Console.WriteLine("-1");
                    return;
                }
            }
            Console.WriteLine(total);
        }
        static T ReadValue<T>()
        {
            return (T)Convert.ChangeType(Console.ReadLine(), typeof(T));
        }
        static (T1, T2) ReadValue<T1, T2>()
        {
            var input = Console.ReadLine().Split(' ');
            return ((T1)Convert.ChangeType(input[0], typeof(T1)), (T2)Convert.ChangeType(input[1], typeof(T2)));
        }
        static (T1, T2, T3) ReadValue<T1, T2, T3>()
        {
            var input = Console.ReadLine().Split(' ');
            return ((T1)Convert.ChangeType(input[0], typeof(T1)), (T2)Convert.ChangeType(input[1], typeof(T2)), (T3)Convert.ChangeType(input[2], typeof(T3)));
        }
        static (T1, T2, T3, T4) ReadValue<T1, T2, T3, T4>()
        {
            var input = Console.ReadLine().Split(' ');
            return ((T1)Convert.ChangeType(input[0], typeof(T1)), (T2)Convert.ChangeType(input[1], typeof(T2)), (T3)Convert.ChangeType(input[2], typeof(T3)), (T4)Convert.ChangeType(input[3], typeof(T4)));
        }
        static T[] ReadValues<T>()
        {
            var input = Console.ReadLine().Split(' ');
            return input.Select(x => (T)Convert.ChangeType(x, typeof(T))).ToArray();
        }
    }
}