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: Accepted)

using System;
using System.IO;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder.Contest.C
{
	static class Program
	{
		public static void Main(string[] args)
		{
			var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
			Console.SetOut(sw);

			var cin = new Scanner();

			int n = cin.NextInt();
			var a = cin.ArrayLong(n + 1);
			long sum = a.Sum();
			long ans = 0;

			long leftnode = 1;
			bool ok = true;
            for (int i = 0; i < n + 1; i++)
            {
				leftnode -= a[i];
				if (leftnode >= 100000000) break;
				if (leftnode < 0)
				{
					ok = false;
					break;
				}
				leftnode *= 2;
            }

			long available = 1;
			if (!ok) Console.WriteLine(-1);
			else
			{
				for (int i = 0; i < n + 1; i++)
				{
					if (available < 1000000000000000L)
					{
						ans += Math.Min(available, sum);
						sum -= a[i];
						available -= a[i];
						available *= 2;
					}
					else
					{
						ans += sum;
						sum -= a[i];
					}
				}

				Console.WriteLine(ans);
			}

			Console.Out.Flush();
		}
	}

	class Scanner
	{
		string[] s;
		int i;

		char[] cs = new char[] { ' ' };

		public Scanner()
		{
			s = new string[0];
			i = 0;
		}

		public string Next()
		{
			if (i < s.Length) return s[i++];
			string st = Console.ReadLine();
			while (st == "") st = Console.ReadLine();
			s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
			if (s.Length == 0) return Next();
			i = 0;
			return s[i++];
		}

		public int NextInt()
		{
			return int.Parse(Next());
		}

		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextInt() + add;
			}
			return Array;
		}

		public long NextLong()
		{
			return long.Parse(Next());
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextLong() + add;
			}
			return Array;
		}

		public double NextDouble()
		{
			return double.Parse(Next());
		}

		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextDouble() + add;
			}
			return Array;
		}
	}
}