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: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] array = new int[n+1];
        for (int i = 0; i < n+1; i++) {
            array[i] = sc.nextInt();
        }
        
        if (n == 0) {
            if (array[0] == 1) {
                System.out.println(1);
                return;
            } else {
                System.out.println(-1);
                return;
            }
        } else {
            if (array[0] != 0) {
                System.out.println(-1);
                return;
            }
        }
        
        long[] sum = new long[n+1];
        sum[n] = array[n];
        for (int i = n-1; i >= 0; i--) {
            sum[i] = sum[i+1] + array[i];
            
        }
        // System.out.println(Arrays.toString(sum));
        
        long[] ans = new long[n+1];
        long now = 1;
        ans[0] = 1;
        now -= array[0];
        if (now < 0) {
            System.out.println(-1);
            return;
        }
        for (int i = 1; i < n+1; i++) {
            now *= 2;
            ans[i] = Math.min(sum[i], now);
            now = ans[i] - array[i];
            if (now < 0) {
                System.out.println(-1);
                return;
            }
        }
        
        long count = 0;
        for (int i = 0; i < n+1; i++) {
            count += ans[i];
        }
        System.out.println(count);
        
    }
}