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;
import java.util.stream.IntStream;

public class Main {
	public static void main(String[] args) {
		var sc = new Scanner(System.in);
		var n = sc.nextInt();
		var a = IntStream.range(0, n + 1).map(i -> sc.nextInt()).toArray();
		sc.close();

		var sum = 0L;
		for (var i = 0; i <= n; i++) {
			sum += a[i];
		}
		var result = 0L;
		var node1 = 1L;
		var node2 = 0L;
		for (var i = 0; i <= n; i++) {
			sum -= a[i];
			var max = i == 0 ? 1 : node1 * 2;
			var node1c = Math.min(max - a[i], sum);
			var node2c = a[i];
			//System.out.printf("%d:%d %d\r\n", i, node1c, node2c);
			if ((node1c + node2c) < node1 || node1c < 0 || i < n && node1c == 0) {
				System.out.println(-1);
				return;
			}
			node1 = node1c;
			node2 = node2c;
			result += (node1 + node2);
		}

		System.out.println(result);
	}
}
