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 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int N = sc.nextInt();
		int[] A = new int[N+1];
		
		for(int i=0; i<=N; i++) {
			A[i] = sc.nextInt();
		}
		
		System.out.println(solve(N, A));
		sc.close();
	}
	
	static long solve(int N, int[] A) {
		if(A[0]>0) {
			return -1;
		}
		
		long sum = 0;
		for(int i=0; i<=N; i++) {
			sum += A[i];
		}
	
		
		long leaf = 1;
		long node = 1;

		for(int i=0; i<N; i++) {
			leaf = Math.min(leaf*2, sum);
			node += leaf;
			if(leaf < A[i+1]) {
				return -1;
			}
			leaf -= A[i+1];
			sum -= A[i+1];
		}
	
		return node;
	}
}