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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt() + 1;
        List<Long> leaf = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            leaf.add(sc.nextLong());
        }
        List<Long> accLeaf = new ArrayList<>();
        accLeaf.add(0L);
        long acc  = 0L;
        for (int i = 0; i < n; i++) {
            acc += leaf.get(i);
            accLeaf.add(acc);
        }

        long  ans = 1;
        long  nonLeaf;
        long  vertexes = 1;
        for (int i = 1; i < n; i++) {
            nonLeaf = vertexes - leaf.get(i-1);
            if (nonLeaf <= 0) {
                System.out.println(-1);
                return;
            }
            vertexes = Math.min(2*nonLeaf, accLeaf.get(n)-accLeaf.get(i));
            ans += vertexes;
        }
        if (vertexes != leaf.get(n-1)) {
            System.out.println(-1);
            return;
        }
        System.out.println(ans);
    }
}
