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.Scanner;

public class Main {
    public static void main(String args[]) {
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();
        int[] a = new int[n + 1];

        int[][] ren = new int[2][n + 1];

        for (int i = 0; i <= n; i++) {
            a[i] = scn.nextInt();
        }
        if (n == 0) {
            if (a[0] == 0) {
                System.out.println(-1);
            } else {
                System.out.println(1);
            }
            return;
        }

        ren[1][n] = a[n];
        ren[0][n] = a[n];
        for (int i = n - 1; i >= 0; i--) {
            ren[0][i] = ren[0][i + 1] + a[i];
            ren[1][i] = a[i] + ((ren[1][i + 1] / 2) + ren[1][i + 1] % 2);
        }
        if (ren[1][0] > 1) {
            System.out.println(-1);
            return;
        }

        int ans = 1;
        int root = 1;

        for (int i = 1; i <= n; i++) {
            root = Math.min((root - a[i-1])*2,ren[0][i]);
            ans += root;
        }
        System.out.println(ans);


        scn.close();


    }


}