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;
using System.Threading.Tasks;

namespace ConsoleApp15
{
    class Class358
    {
        static void Main()
        {
            int N = int.Parse(Console.ReadLine());
            var A = Console.ReadLine().Split().Select(long.Parse).ToArray();

            if (A[0] != 0) { Console.WriteLine(-1);return; }
            long[] max = new long[N+1];
            max[N] = A[N];
            long[] mi = new long[N + 1];
            mi[N] = 0;
            
            for(int i = N - 1; i >= 0; i--)
            {
                var a = A[i];
                mi[i] =mi[i+1] +  (max[i+1] - mi[i+1]) / 2;
                max[i] = max[i+1] + a;
                
            }
            /*
            if (max[0] - mi[0] > 1) { Console.WriteLine(-1);return; }
            for(int i = 0; i < N+1; i++)
            {
                Console.WriteLine( $"{max[i]} {mi[i]}");
            }
            */
            long sum = 1;
            long leaf = 1;
            for(int i = 1; i < N + 1; i++)
            {
                leaf = Math.Min((leaf - A[i-1]) * 2, max[i]);
                sum += leaf;
            }
            Console.WriteLine(sum);
        }
    }
}