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 {

    Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Main().run();
    }

    void run() {

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

        long[] max = new long[n+1];
        max[0] = 1 - arr[0];
        if (max[0] < 0) {
            System.out.println(-1);
            return;
        }

        long INF = (long)1e14;
        for (int i=1; i<n+1; i++) {
            if (max[i-1]*2 - arr[i] < 0) {
                System.out.println(-1);
                return;
            }
            if (max[i-1]*2 - arr[i] > INF) {
                for (int j=i; j<n+1; j++) {
                    max[j] = INF;
                }
                break;
            } else {
                max[i] = max[i-1]*2 - arr[i];
            }
        }

//        System.out.println(max[n]);

        long sum = 0;
        long ans = 0;
        for (int i=n; 0<=i; i--) {
            sum = Math.min(sum, max[i]);
            ans += (sum+arr[i]);
            sum += arr[i];
//            System.out.println(sum);
        }

        System.out.println(ans);

    }
}