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;

class Main {
    final Scanner sc = new Scanner(System.in);

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

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

        // 葉ではない頂点の最大数
        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;
            }
        }

        long ans = 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];
            ans += b[i];
        }

        System.out.println(ans);
    }
}

