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.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner scan = new Scanner(System.in);
        TreeMap<Integer, Integer> tm = new TreeMap<Integer, Integer>();
        int n= Integer.parseInt(scan.next());
        int cnt=0;
        for(int i=0;i<n+1;i++){
            int wk= Integer.parseInt(scan.next());
            if(tm.get(wk)!=null ){
                for(int j=wk;true;j++){
                    if(tm.get(j)==null){
                        for(int k=0;k<2;k++,cnt++){
                            tm.put(j,cnt);
                        }
                        break;
                    }
                }
            }else{
                for(int k=0;k<2;k++,cnt++){
                    tm.put(wk,cnt);
                }
            }
            
            
            
        }
        if(tm.get(n)==null){
            System.out.println(-1);
        }
        System.out.println(tm.get(n));
    }
}