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();
        int[] a = new int[n+1];
        for(int i=0; i<=n; i++) {
            a[i] = sc.nextInt();
        }

        long res = 0;
        long[] max = new long[n+1];
        max[0] = 1 - a[0];
        if(max[0] < 0) {
            System.out.println(-1);
            return;
        }
        long ln = 100000000000000L;
        for(int i=1; i<=n; i++) {
            max[i] = Math.min(max[i-1]*2 - a[i], ln);
            if(max[i] < 0) {
                System.out.println(-1);
                return;
            }
        }

        res = a[n];
        long[] b = new long[n+1];
        b[n] = a[n];
        for(int i=n-1; i>=0; i--) {
            b[i] = Math.min(max[i], b[i+1]) + a[i];
            res += b[i];
        }

        System.out.println(res);
    }
}