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[] arr = new int[n + 1];
		for (int i = 0; i <= n; i++) {
		    arr[i] = sc.nextInt();
		}
		if (n == 0) {
		    if (arr[0] == 1) {
		        System.out.println(1);
		    } else {
		        System.out.println(-1);
		    }
		    return;
		} else {
		    if (arr[0] != 0) {
		        System.out.println(-1);
		        return;
		    }
		}
		long[] sums = new long[n + 1];
		sums[n] = arr[n];
		for (int i = n - 1; i >= 0; i--) {
		    sums[i] = sums[i + 1] + arr[i];
		}
		long total = 0;
		long current = 1;
		for (int i = 0; i < n; i++) {
		    total += current;
		    current -= arr[i];
		    if (current <= 0) {
		        System.out.println(-1);
		        return;
		    }
		    current = Math.min(current * 2, sums[i + 1]);
		}
		if (current != arr[n]) {
		    System.out.println(-1);
		    return;
		}
		total += current;
		System.out.println(total);
	}
}
