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.Scanner;

public class Main {
    static public void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt()+1;
        long[] A = new long[N];
        long branch = 0;
        for(int i = 0 ; i < N ; i++){
            A[i] = sc.nextLong();
            branch += A[i];
        }
        branch--;

        long ans = 0;
        long nowLeaf = 1;
        for(int i = 0 ; i < N ; i ++){
            ans += nowLeaf;
            nowLeaf -= A[i];
            if(nowLeaf < 0){
                ans = -1;
                break;
            }
            if(nowLeaf <= branch){
                branch -= nowLeaf;
                nowLeaf *= 2;
            }else{
                nowLeaf += branch;
                branch = 0;
            }
        }

        System.out.println(ans);
    }
}