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) {

		Scanner sc = new Scanner(System.in);
		
		int n = Integer.parseInt(sc.next());
		
		long []a = new long [n+1];
		long node = 0L;
		
		for(int i = 0 ; i <= n ;i++) {
			a[i] = Long.parseLong(sc.next());
			node += a[i];
		}
		
		long []b = new long[n+1];
		b[0] = 1;
		long sum = b[0];
		
		for(int i = 0 ; i < n ;i++) {
			if(b[i] <= a[i]) {
				System.out.println(-1);
				return;
			}
			
			node -= a[i];
			b[i+1] = Math.min(node, (b[i] - a[i])*2);
			sum += b[i+1];
			
		}
		
		if(b[n] < a[n]) {
			System.out.println(-1);
          	return;
		}
		
		System.out.println(sum);
		
	}
}
