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);
		long sum = 0;
		int N = sc.nextInt();
		long A[] = new long[N+1];
		long result[] = new long[N+1];
		for (int i=0;i<=N;i++) {
			A[i] = sc.nextLong();
		}
		result[N]= A[N];
		for (int i=N;i>=1;i--) {
			result[i-1]= A[i-1]+result[i];
		}
		long tempPoint=1;
		for (int i=0;i<=N;i++) {
			if (tempPoint<A[i]) {
				System.out.println(-1);
				return ;
			}
			if (tempPoint<result[i]) {
				sum +=tempPoint;
			} else {
				sum +=result[i];
			}
			tempPoint-=A[i];
			if (tempPoint<Long.MAX_VALUE/4) {
				tempPoint*=2;
			}
		}
		System.out.println(sum);
	}
}