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.util.function.*;
import java.math.*;
public class Main {
    public static void main(String[] args){
		Scanner s = new Scanner(System.in);
		int N = s.nextInt();
		long[] leaf = new long[N + 1];
		for(int i = 0; i < N + 1; i++) {
			leaf[i] = s.nextInt();
		}
		long[] MAX = new long[N+2];
		long[] MIN = new long[N+2];
		MAX[N] = MIN[N] = leaf[N];
		
		for(int i = N - 1; i >= 0; i--) {
			MAX[i] = MAX[i+1] + leaf[i];
			MIN[i] = (MIN[i+1]+1) / 2 + leaf[i]; 
		}
		
		if(MIN[0] != 1) {
			System.out.println("-1");
			return;
		}
		
		long cur = 1;
		long ret = 0;
		for(int i = 0; i <= N; i++) {
			ret += cur;
			cur -= leaf[i];
			cur = Long.min(cur*2, MAX[i+1]);
		}
		System.out.println(ret);
	}
}