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: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	static int N;
	static int[] A;
	static int[] B;
	static int res = -1;
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		read();
	}

	private static void read() throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		N = Integer.parseInt(br.readLine());
		String[] words = br.readLine().split(" ");
		A = new int[N+1];
		B = new int[N+1];
		for(int i = 0;i<N+1;i++){
			A[i] =  Integer.valueOf(words[i]);
		}
		search(0,0,0);
		System.out.println(res);
		br.close();
	}
	
	public static void search(int depth,int sum,int lastB) {
		if ( depth == N+1){
			res = sum;
			return;
		}
		
		if( res != -1){
			return;
		}
		
		int a = A[depth];
		int[] range = rangeB(depth, lastB);
		int maxB =range[0];
		int minB = range[1];
		
		if( maxB < 0){
			return;
		}
		
		if( depth == N) {
			if ( minB<=0 && maxB>=0){
				search(depth+1,sum+a+0,0);
			}
		}
		else{
			for( int b = maxB; b>=minB;b--){
				B[depth] = b;
				search(depth+1,sum+a+b,b);
			}
		}
	}
	
	public static  int[] rangeB(int depth,int lastB ){
		if( depth == 0 ){
			if ( A[depth]> 1){
				return new int[]{-1,-1};
			}
			else if ( A[depth] == 1 ){
				return new int[]{0,0};
			}
			return new int[]{1,1};
		}
		int maxB = 2*lastB - A[depth];
		
		if (maxB  < 0 ){
			return new int[]{-1,-1};
		}
	
		int minB = lastB - A[depth];
		if ( minB < 0 ){
			minB = 0;
		}
		return new int[]{maxB,minB};
	}
}