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.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[] a = new long[n+1];
        for (int i = 0; i < n+1; i++) {
            a[i] = sc.nextLong();
        }
        if(a[0] != 0){
            if(n == 0 && a[0] == 1) System.out.println(1);
            else System.out.println(-1);
        }else{
            long[] max = new long[n+1];
            max[n] = a[n];
            for (int i = n; i > 0; i--) {
                max[i-1] = max[i] + a[i-1];
            }
            max[0] = 1;
            long ans = 1L;
            for (int i = 0; i < n; i++) {
                max[i+1] = Math.min(max[i+1], (max[i] - a[i]) * 2);
                ans += max[i+1];
                if(max[i+1] < a[i+1]){
                    ans = -1;break;
                }
            }
            System.out.println(ans);
        }
        sc.close();

    }

}
