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.*;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		int n = Integer.parseInt(input.nextLine());
		
		String[] values = input.nextLine().split(" ");
	    int[] val = new int[values.length];
	    
	    for(int v=0; v< values.length; v++) {
	    	val[v] = Integer.parseInt(values[v]);
	    }
	    
	    int count =1;
	    int leaves = 1;
	    int maxleaves = 0;
	    int usedLeaves = leaves;
	   if(leaves <2) 
	    for (int i=0; i<n+1; i++) {
	    
	    	 maxleaves= 2*usedLeaves;
	    	 if(val[i] <= maxleaves) {
	    		 usedLeaves = maxleaves - (val[i]*2);
		    	 count = count + usedLeaves;
	    	 } else {
	    		 count =-1;
	    		 break;
	    	 }
	    }
	   
	   System.out.println(count);
	  
	}
}