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: Runtime Error)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();

        long ans = 0;

        int array[] = new int[n + 1];

        for (int i = 0; i < n + 1; i++) {
            array[i] = sc.nextInt();
        }

        int max[] = new int[n];
        max[0] = 1;
        //チェックと計算
        for (int i = 0; i < n; i++) {
            if (array[i] > max[i] || array[0] > 0) {
                System.out.println(-1);
                return;
            }

            if (i == n - 1) {
                break;
            }

            max[i + 1] = max[i] * 2 - array[i + 1];
        }

        int sum = 0;

        for (int i = 0; i < n + 1; i++) {

            if (i == 0) {
                ans += array[n];
                sum += array[n];
                continue;
            }

            if (max[n - i] >= sum + array[n - i]) {
                sum += array[n - i];
                ans += sum;
            } else{
                ans += max[n - i] + array[n - i];
            }
        }

        System.out.println(ans);

        sc.close();
    }

}