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.Runtime.InteropServices;

public class Solution
{
    public static void Main()
    {
        var N = int.Parse(Console.ReadLine()) + 1;
        var A = Console.ReadLine().Split(' ').Select(long.Parse).ToArray();

        if (N == 0 && A[0] == 0)
        {
            Console.WriteLine(-1);
            return;
        }

        var max = new long[N];
        max[N - 1] = A[N - 1];
        for (int i = N - 2; i >= 0; i--)
        {
            max[i] = max[i + 1] + A[i];
        }

        long count = 0;
        long cur = 1;
        for (int i = 0; i < N - 1; i++)
        {
            if (cur <= A[i])
            {
                Console.WriteLine(-1);
                return;
            }

            count += cur;
            var next = Math.Min(max[i + 1], (cur - A[i]) * 2);
            cur = next;
        }
        count += A.Last();

        Console.WriteLine(count);
    }
}