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

namespace debug
{
    class main
    {
        static void Main(string[] args)
        {
            //問題クラスを展開
            ProgramC a = new ProgramC();
            a.main();//実行する

        }

    }

    class ProgramA
    {
        public void main()
        {

            string[] s = Console.ReadLine().Split(' ');
            int a = int.Parse(s[0]);
            int b = int.Parse(s[1]);
            int c = int.Parse(s[2]);
            int d = int.Parse(s[3]);
            int e = int.Parse(s[4]);

            Console.WriteLine(d + c * 60 - b - a * 60 - e);

        }
    }

    class ProgramB
    {
        public void main()
        {

            string s = Console.ReadLine();

            for(int i = 0;i < s.Length;i++)
            {
                if(s[i] == '?')
                {
                    Console.Write('D');
                }
                else
                {
                    Console.Write(s[i]);
                }
            }
            

        }
    }

    class ProgramC
    {
        public void main()
        {

            int n = int.Parse(Console.ReadLine());
            string[] a = Console.ReadLine().Split(' ');

            //n=0のときは例外処理
            /*if(n == 0)
            {
                if (a[0] == "1")
                    Console.WriteLine("1");
                else
                    Console.WriteLine("-1");
                return;
            }*/
            //0のときは0以外はダメ
            if(a[0] != "0")
            {
                Console.WriteLine("-1");
                return;
            }
            long[] num = new long[n + 1];
            long[] leave = new long[n + 1];
            num[0] = 1;
            num[n] = long.Parse(a[n]);
            leave[n] = long.Parse(a[n]);
            //深さiを含む後ろの根の数を数えておく
            for (int i = n -1; i >= 0 ;i--)
                leave[i] = leave[i + 1] + long.Parse(a[i]);
            //前から根の数を決める（ただし、最大は後ろに考えるべき根の数に限られる）
            for(int i = 1;i < n;i++)
            {
                num[i] = Math.Min(2 * (num[i - 1] - long.Parse(a[i - 1])), leave[i]);
                //あり得る数が葉になる数より大きければあり得ない
                if(num[i] <= long.Parse(a[i]))
                {
                    Console.WriteLine("-1");
                    return;
                }
            }
            
            //答え出力
            long ans = 0;
            for(int i = 0; i <= n;i++)
                ans += num[i];

            Console.WriteLine(ans);
        }
    }
}