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;

public class AtCoderA
{
    static void Main()
    {
        var sc = new Scanner();

        int n = sc.NextInt();
        int[] a = sc.IntArray();
        int[] afterSum = new int[n + 1];
        for (int i = n; i >= 0; i--)
        {
            if (i == n)
            {
                afterSum[i] = a[i];
            }
            else
            {
                afterSum[i] = a[i] + afterSum[i + 1];
            }
        }

        long answer = 1;
        long current = 1 - a[0];

        if (current < 0)
        {
            Console.WriteLine(-1);
            return;
        }

        for (int i = 1; i <= n; i++)
        {
            current *= 2;
            if (a[i] > current)
            {
                Console.WriteLine(-1);
                return;
            }
            current = Math.Min(afterSum[i], current);
            answer += current;
            current -= a[i];
            if (current < 0)
            {
                Console.WriteLine(-1);
                return;
            }
            if (i == n && current != 0)
            {
                Console.WriteLine(-1);
                return;
            }
        }

        Console.WriteLine(answer);
    }

    class Scanner
    {
        private string[] _line;
        private int _index;
        private const char Separator = ' ';

        public Scanner()
        {
            _line = new string[0];
            _index = 0;
        }

        public string Next()
        {
            while (_index >= _line.Length)
            {
                _line = Console.ReadLine().Split(Separator);
                _index = 0;
            }

            return _line[_index++];
        }

        public int NextInt() => int.Parse(Next());
        public long NextLong() => long.Parse(Next());
        public double NextDouble() => double.Parse(Next());
        public decimal NextDecimal() => decimal.Parse(Next());
        public char NextChar() => Next()[0];
        public char[] NextCharArray() => Next().ToCharArray();

        public string[] Array()
        {
            _line = Console.ReadLine().Split(Separator);
            _index = _line.Length;
            return _line;
        }

        public int[] IntArray() => Array().Select(int.Parse).ToArray();
        public long[] LongArray() => Array().Select(long.Parse).ToArray();
        public double[] DoubleArray() => Array().Select(double.Parse).ToArray();
        public decimal[] DecimalArray() => Array().Select(decimal.Parse).ToArray();
    }
}