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.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.ListIterator;
import java.util.Map.Entry;


import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;


public class Main {
	
	public static void main(String[] args) throws IOException {	
		try(final Scanner sc = new Scanner(System.in)){
			final int N = sc.nextInt();
			
			long[] as = new long[N + 1];
			long sum = 0;
			for(int i = 0; i <= N; i++) {
				as[i] = sc.nextLong();
				sum += as[i];
			}
			
			long current_nodes = 1;
			long answer = 1;
			for(int i = 0; i <= N; i++) {
				//System.out.println(free + " " + as[i]);

				current_nodes -= as[i];
				sum -= as[i];
				if(current_nodes < 0) {
					System.out.println(-1);
					return;
				}
				
				final long next_nodes = Math.min(current_nodes * 2, sum);
				//System.out.println(free + "- next -> " + next_free);
				current_nodes = next_nodes;
				answer += current_nodes;
			}
			
			System.out.println(answer);
		}
	}

	public static class Scanner implements Closeable {
		private BufferedReader br;
		private StringTokenizer tok;

		public Scanner(InputStream is) throws IOException {
			br = new BufferedReader(new InputStreamReader(is));
		}

		private void getLine() throws IOException {
			while(!hasNext()){
				tok = new StringTokenizer(br.readLine());
			}
		}

		private boolean hasNext() {
			return tok != null && tok.hasMoreTokens();
		}

		public String next() throws IOException {
			getLine();
			return tok.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
		

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public int[] nextIntArray(int n) throws IOException {
			final int[] ret = new int[n];
			for(int i = 0; i < n; i++){
				ret[i] = this.nextInt();
			}
			return ret;
		}

		public long[] nextLongArray(int n) throws IOException {
			final long[] ret = new long[n];
			for(int i = 0; i < n; i++){
				ret[i] = this.nextLong();
			}
			return ret;
		}

		public void close() throws IOException {
			br.close();
		}
	}
}