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;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n + 1];
		for (int i = 0; i < a.length; i++) {
			a[i] = sc.nextInt();
		}
		sc.close();

		long inf = 1000000000000000L;
		long[] max = new long[n + 1];
		max[0] = 1 - a[0];
		if (max[0] < 0) {
			System.out.println(-1);
			return;
		}
		for (int i = 1; i <= n; i++) {
			max[i] = max[i - 1] * 2 - a[i];
			if (max[i] < 0) {
				System.out.println(-1);
				return;
			}
			if (max[i] >= inf) {
				for (int j = i; j <= n; j++) {
					max[j] = inf;
				}
				break;
			}
		}

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