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 stdIn=new Scanner(System.in);
		int N=stdIn.nextInt();
		int A[]=new int[N+1];
		long key[]=new long[N+1];
		long ans=0,sum=1,cout=0;
		for(int i=0;i<N+1;i++){
			A[i]=stdIn.nextInt();
		}
		ans=A[N];key[N]=A[N];
		for(int i=N-1;i>=0;i--){
			key[i]=key[i+1]+A[i];
			ans+=key[i];
		}
		for(int i=0;i<N+1;i++){
			if(i!=N&&A[i]+1>sum){
				System.out.println(-1);
				System.exit(0);
			}
			if(i==N&&A[i]>sum){
				System.out.println(-1);
				System.exit(0);
			}
			else
				if(key[i]>sum){
					ans-=key[i]-sum;key[i]=sum;
				}
			sum=(key[i]-A[i])*2;
		}
		System.out.println(ans);
	}
}
