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.Arrays;
import java.util.Scanner;

class C {
    public static void main(String... args) {
        final Scanner sc = new Scanner(System.in);
        final int N = sc.nextInt();
        final int[] A = new int[N + 1];
        Arrays.setAll(A, $ -> sc.nextInt());
        if (N == 0) {
            System.out.println(A[0] == 1 ? 1 : -1);
            return;
        }
        if (!possible(A, N)) {
            System.out.println(-1);
            return;
        }
        int lo = 0;
        int up = N;
        while (up - lo > 1) {
            final int x = (lo + up + 1) / 2;
            if (possible(A, x))
                up = x;
            else
                lo = x;
        }
        final int x = (lo + up + 1) / 2;
        System.err.println(x);
        long ans = 0;
        long nodes = 0;
        for (int i = N; i >= x; i--) {
            nodes += A[i];
            ans += nodes;
        }
        for (int i = x - 1; i >= 0; i--) {
            nodes = (nodes + 1) / 2;
            nodes += A[i];
            ans += nodes;
        }
        System.out.println(ans);
    }

    static boolean possible(final int[] A, final int p) {
        long lo = 0;
        for (int i = p; i < A.length; i++)
            lo += A[i];
        long up = 1;
        for (int i = 0; i < p; i++) {
            up -= A[i];
            up *= 2;
        }
        return up >= lo;
    }
}
public class Main {
    public static void main(String... args) {
        C.main();
    }
}