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;

class Program
{
	static void Main(string[] args)
	{
		string[] input = Console.ReadLine().Split(' ');
		int n = int.Parse(input[0]);
		int[] a = new int[n+1];
		input = Console.ReadLine().Split(' ');
		long leaves = 0;
		for(int i = 0;i <= n; i++)
		{
			a[i] = int.Parse(input[i]);
			leaves += (long)a[i];
		}

		long[] point = new long[n+1];
		point[0] = 1;
		long allPoints = point[0];
		for(int i = 0;i < n; i++)
		{
			long notLeaf = point[i] - (long)a[i];
			leaves -= (long)a[i];
			if(2*notLeaf<(long)a[i+1])
			{
				Console.WriteLine(-1);
				return;
			}
			point[i+1] = (int)Math.Min(2*notLeaf, leaves);
			allPoints += point[i+1];
		}
		if(point[n]!=a[n]) allPoints = -1;
		Console.WriteLine(allPoints);

	}

}