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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {
	public static void main(String[] args) {
		var sc = new Scanner(System.in);
		var n = sc.nextInt();
		var a = IntStream.range(0, n + 1).map(i -> sc.nextInt()).toArray();
		sc.close();

		var sum = Arrays.stream(a).sum();
		var result = 0L;
		var node1 = 1;
		var node2 = 0;
		for (var i = 0; i <= n; i++) {
			sum -= a[i];
			var max = i == 0 ? 1 : node1 * 2;
			var node1c = Math.min(max - a[i], sum);
			var node2c = a[i];
			if ((node1c + node2c) < node1 || i < n && node1c <= 0) {
				System.out.println(-1);
				return;
			}
			node1 = node1c;
			node2 = node2c;
			result += (node1 + node2);
		}

		System.out.println(result);
	}
}