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;

namespace AtCoder3005
{
	class Program
	{
		static void Main(string[] args)
		{
			C();
		}

		static void C()
		{
			var n = int.Parse(Console.ReadLine());
			var a = Console.ReadLine().Split().Select(x => long.Parse(x)).ToList();
			if (n > 0 && a[0] != 0)
			{
				Console.WriteLine(-1);
				return;
			}
			if (n == 0 && a[0] == 1)
			{
				Console.WriteLine(1);
				return;
			}
			if (n == 0 && a[0] != 1)
			{
				Console.WriteLine(-1);
				return;
			}

			var maxAtLevelsByTop = new long[n + 1];
			var maxAtLevelsByBot = new long[n + 1];
			var minAtLevelsByBot = new long[n + 1];
			minAtLevelsByBot[n] = a[n];
			maxAtLevelsByBot[n] = a[n];

			for (int i = n - 1; i >= 0; i--)
			{
				var leavesAtLevel = a[i];
				var minNext = minAtLevelsByBot[i + 1];
				var maxNext = maxAtLevelsByBot[i + 1];

				var currentMin = leavesAtLevel + (minNext + 1 )/ 2;
				var currentMax = leavesAtLevel + maxNext;
				minAtLevelsByBot[i] = currentMin;
				maxAtLevelsByBot[i] = currentMax;
			}

			maxAtLevelsByTop[0] = 1;

			for (int i = 1; i <= n; i++)
			{
				var lastMax = maxAtLevelsByTop[i - 1];
				var possible = (lastMax - a[i - 1]) * 2;
				maxAtLevelsByTop[i] = possible;
			}

			long count = 0;
			for (int i = 0; i <= n; i++)
			{
				if (maxAtLevelsByTop[i] < minAtLevelsByBot[i])
				{
					Console.WriteLine(-1);
					return;
				}

				var maxCount = Math.Min(maxAtLevelsByTop[i], maxAtLevelsByBot[i]);
				count += maxCount;
			}

			Console.WriteLine(count);
		}

		static void B()
		{
			var str = Console.ReadLine();
			var newStr = str.Replace('?', 'D');
			Console.WriteLine(newStr);
		}

		static void A()
		{
			var ints = Console.ReadLine().Split().Select(x => int.Parse(x)).ToList();
			var h1 = ints[0];
			var m1 = ints[1];
			var h2 = ints[2];
			var m2 = ints[3];
			var k = ints[4];

			var start = 60 * h1 + m1;
			var end = 60 * h2 + m2;
			if (end < start)
				end += 1440;

			var diff = end - start;
			Console.WriteLine(diff - k);
		}
	}
}