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 {
    private static final long INF =  1_000_000_000_000_000_000L;
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] A = new int[N+1];
        for (int i = 0; i <= N; i++) {
            A[i] = sc.nextInt();
        }

        long[] pow = new long[Math.max(N+1,32)];
        for (int i = 0; i < N+1; i++) {
            pow[i] = INF;
        }
        pow[0] = 1;
        for (int i = 1; i <= 30; i++) {
            pow[i] = 2*pow[i-1];
        }
        // pow[30] 1_000_000_000;

        long[] PMax = new long[N+1];
        for (int i = 0; i < N+1; i++) {
            PMax[i] = INF;
        }

        PMax[0] = 1;
        for (int i = 1; i <= N-1; i++) {
            PMax[i] = Math.min(2*PMax[i-1]-A[i],INF);
            if (PMax[i] == INF) {
                break;
            }
            if (PMax[i] <= 0) {
                System.out.println(-1);
                return;
            }
        }

        if (A[N] > pow[N]) {
            System.out.println(-1);
            return;
        }
        long ans = A[N];
        long pre = A[N];
        for (int i = N-1; i >= 0; i--) {
            long PMax2 = Math.min(PMax[i],pre);
            long now = PMax2+A[i];
            if (now > pow[i]) {
                System.out.println(-1);
                return;
            }
            ans += now;
            pre = now;
        }

        System.out.println(ans);
    }
}