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: Accepted)

import java.io.*;
import java.util.*;
public class Main {
	
	public static void main(String... args) {
		FastScanner sc = new FastScanner();
		int n = sc.nextInt();
		long[] arr = new long[n+1];
		for(int i = 0; i <= n; i++) {
			arr[i] = sc.nextInt();
		}
		if(n == 0) {
			if(arr[0] == 1) {
				System.out.println(1);
			}
			else {
				System.out.println(-1);
			}
			return;
		}
//		if(arr[0] > 0) {
//			System.out.println(-1); return;
//		}
		long[] nl = new long[n+1]; //not leaves
		for(int i = n-1; i >= 0; i--) {
			nl[i] = nl[i+1] + arr[i+1];
		}
		long max = 1;
		long res = 0;
		for(int i = 0; i <= n; i++) {
			if(nl[i] + arr[i] > max) {
				nl[i] = max - arr[i];
				if(nl[i] < 0) {
					System.out.println(-1); return;
				}
				max = 2 * nl[i];
			}
			res += nl[i] + arr[i];
		}
		System.out.println(res);
	}
	
	static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }
	

}