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 class abc167_a{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
		
		int n = scan.nextInt();
		long[] arr = new long[n+1]; 
		for(int i=0; i<= n; i++){
			arr[i] = scan.nextLong();
		}
		long num = 0;
		long eda = 1;
		boolean flag = true;
		for(int i=0; i<= n; i++){
			if(i == 0 && arr[i] > 0){
				flag = false;
              	System.out.println("yes1"+i);
				break;
			}
			if( !(0 <= arr[i] || arr[i] < eda) ){
				if(arr[i] == eda && i== n){
					num += eda;
					break;
				}
				flag = false;
              System.out.println("yes2"+i);
				break;
			}

			num += eda;
			eda = (eda - arr[i])*2;

		}
		if(flag){
			System.out.println(num);
		}else{
			System.out.println(-1);
		}
	}

}