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.Scanner;
public class Main {
	public static void main(String[] args)throws Exception{
		Scanner stdIn=new Scanner(System.in);
		int N=stdIn.nextInt();
		int A[]=new int[N+1];
		long key[]=new long[N+1];
		long next=1,ans=0,cout=0;
		for(int i=0;i<N+1;i++){
			A[i]=stdIn.nextInt();
			key[i]=next;
			ans+=next;
			next-=A[i];
			if(next<0){
				System.out.println(-1);
				System.exit(0);
			}
			next*=2;
		}
		cout=A[N];ans-=key[N]-A[N];
		for(int i=N-1;i>=0;i--){
			if(key[i]>=cout+A[i]){
				ans-=key[i]-(cout+A[i]);
				if(A[i]>0)
					cout+=A[i];
			}
			else
				cout=key[i];
		}
		System.out.println(ans);
	}
}