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.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int[] a=new int[n+1];
		long[] vol=new long[n+1];
		long two=1;
		vol[0]=1;
		for(int i=0; i<=n; i++){
			a[i]=sc.nextInt();
			if(a[i]>vol[i]){
				System.out.println(-1);
				return;
			}
			if(i!=n){
				vol[i+1]=Math.min(10000000000000L,two=(two-a[i])*2);
				two=Math.min(10000000000000L,two);
			}
		}
		long ans=0;
		long num=0;
		for(int i=n; i>=0; i--){
			num+=a[i];
			num=Math.min(vol[i],num);
			ans+=num;
		}
		System.out.println(ans);
	}
}
