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 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main{
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt() + 1;
		long[] A = new long[N];
		long ans = 1;
		long[] tyoten = new long[N];
		for(int i = 0;i < N;i++){
			A[i] = sc.nextLong();
		}
		tyoten[0] = 1;
		if(A[0] > 0){
			tyoten[0]--;
		}
		if(N == 1 && A[0] != 1){
			ans = -1;
		}
		for(int i = 1;i < N;i++){
			tyoten[i] = tyoten[i-1]*2;
			ans += tyoten[i];
			tyoten[i] -= A[i];
			if(tyoten[i] < 0){
				ans = -1;
				break;
			}
		}
		if(ans != -1 && N > 1){
			long[] tyo = new long[N];
			ans -= tyoten[N-1];
			if(A[N-1] < tyoten[N-2]){
				tyo[N-2] = A[N-1];
			}else{
				tyo[N-2] = tyoten[N-2];
			}
			for(int i = N - 2;i > 0;i--){
				ans -= tyoten[i];
				ans += tyo[i];
				if(A[i] + tyo[i] < tyoten[i-1]){
					tyo[i-1] = A[i] + tyo[i];
				}else{
					tyo[i-1] = tyoten[i-1];
				}
			}
		}
		System.out.println(ans);
	}
}