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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {
		Scanner s = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		String[] input;
		int n = Integer.parseInt(reader.readLine());
		long[] arr = new long[n + 1];
		input = reader.readLine().split(" ");
		for(int i = 0; i <= n; i++) {
			arr[i] = Integer.parseInt(input[i]);
		}
//		if(n > 0 && (arr[0] != 0 || arr[n] == 0)) {
//			System.out.println(-1);
//			return;
//		}
		long[][] up = new long[n + 1][2];
		up[0][0] = 1;
		for(int i = 1; i <= n; i++) {
			up[i][0] = ((up[i - 1][0]) * 2) - arr[i];
			if(up[i - 1][0] > Math.pow(10, 18)) {
				up[i][0] = Long.MAX_VALUE;
			}
			if(up[i][0] < 0) {
				System.out.println(-1);
				return;
			}
//			System.out.println(up[i][0] + " ");
		}
		up[n][1] = 0;
		for(int i = n - 1; i >= 0; i--) {
			up[i][1] = arr[i + 1] + up[i + 1][1];
		}
		BigInteger ans = new BigInteger("0");
		long[] temp = new long[n + 1];
		for(int i = 0; i <= n; i++) {
			ans = ans.add(new BigInteger((Math.min(up[i][0], up[i][1]) + arr[i]) + ""));
			temp[i] = Math.min(up[i][0], up[i][1]);
			if(temp[i] < 0) {
				System.out.println(-1);
				return;
			}
		}
		if(Long.parseLong(ans.toString()) > 1 && arr[0] != 0) {
			System.out.println(-1);
			return;
		}
		out.print(ans.toString());
		out.flush();
	}

}
