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.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
      	Integer A[] = new Integer[N+1];
        for (int i=0; i<=N; i++) {
            A[i] = sc.nextInt();
        }
      	int ans = 0;
      	int d = 1;
      	for(int j=1; j<N; j++){
        if(d <= 0) break;
          if(A[j] == 0){
            if(d * 2 <= A[N]){
              ans = ans + d*2;
              d = d*2;
            }
            else{
              ans = ans + A[N];
              d = A[N];
            }
          }
          else {
            if(d * 2 - A[j] <= A[N]){
              ans = ans + d*2 - A[j];
              d = d*2 - A[j];
            }
            else{
              ans = ans + A[N];
              d = A[N];
            }
          }
        //System.out.println(ans);  
        //System.out.println("d="+d);
        }
      	int sum = 0;;
        for( int i=0; i<A.length; i++ ) {
          sum += A[i];
        }
      	//System.out.println(sum);
        ans = ans + sum + 1;
      	if(d>=2){
        System.out.println(ans); }
      	else{
          System.out.println("-1");
   		 }
    }
}