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[]) {

		Scanner sc = new Scanner(System.in);
		
		int N = sc.nextInt();

		long result = 0;
		long[] A = new long[N + 1];
		for(int i = 0; i < N + 1; i++) {
			A[i] = sc.nextLong();
			result += A[i];
		}

		long[] B = new long[N + 1];
		long[] C = new long[N + 1];
		
		long[] limit = new long[N + 1];
		Arrays.fill(limit, Long.MAX_VALUE);
		limit[0] = 1;

		for(int i = 1; i < N + 1; i++) {
			if(limit[i - 1] < Long.MAX_VALUE / 2)limit[i] = limit[i - 1] * 2;
			else break;
		}
		
		boolean flag = true;
		if(A[0] == 0)B[0] = 1;
		else if(A[0] == 1) B[0] = 0;
		else flag = false;
		for(int i = 1; i < N + 1; i++) {
			B[i] = B[i - 1] * 2 - A[i];
			if(B[i] < 0) {
				flag = false;
				break;
			}
		}
		
		
		if(flag) {
			C[N] = 0;
			for(int i = N - 1; i >= 0; i--) {
				C[i] = Math.min(C[i + 1] + A[i + 1], limit[i] - A[i]);
			}
			
			for(int i = 0; i < N + 1; i++) {
				result += Math.min(B[i], C[i]);
			}
			
		}
		
		System.out.println((flag)?result:-1);
		
		
		sc.close();
	}
}