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: Runtime Error)

using System;
using System.Linq;

class C
{
	static void Main()
	{
		var n = int.Parse(Console.ReadLine());
		var a = Console.ReadLine().Split().Select(int.Parse).ToArray();

		var p = a.Select((x, i) => (1L << Math.Min(60, i)) - x).ToArray();
		if (p.Where((x, i) => i < n ? x <= 0 : x < 0).Any()) { Console.WriteLine(-1); return; }

		p[n] = 0;
		for (int i = n - 1; i >= 0; i--)
			p[i] = Math.Min(p[i], p[i + 1] + a[i + 1]);
		for (int i = 1; i <= n; i++)
			p[i] = Math.Min(p[i], 2 * p[i - 1] - a[i]);
		Console.WriteLine(a.Sum() + p.Sum());
	}
}