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.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] sp = br.readLine().split(" ");
        long[] a = new long[n+1];
        for(int i = 0; i <= n; i++) {
            a[i] = Long.parseLong(sp[i]);
        }

        long[] upL = new long[n+1];
        long[] upR = new long[n+1];
        long[] down = new long[n+1];
        
        upL[n] = a[n];
        upR[n] = a[n];
        for(int i = n-1; i >= 0; i--) {
            upL[i] = upL[i+1]/2 + a[i];
            if(upL[i+1]%2 != 0) upL[i]++;
            upR[i] = upR[i+1] + a[i];
        }

        if(upL[0] != 1) {
            System.out.println(-1);
            return;
        }
        
        down[0] = 1;
        for(int i = 1; i <= n; i++) {
            down[i] = (down[i-1] - a[i-1]) * 2;
        }
        long ans = 0;
        for(int i = 0; i <= n; i++) {
            ans += Math.min(upR[i], down[i]);
        }
        System.out.println(ans);
    }
}