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 sc = new Scanner(System.in);
		long N = sc.nextInt();
		long[] input = new long[(int) (N + 1)];
		long div = 0;
		for (int i = 0; i < N + 1; i++) {
			input[i] = sc.nextInt();
			div += input[i];
		}
		div--;
		long nowTop = 1 - input[0];
		long numOfTop = 1;

		for (int i = 0; i < N; i++) {
			// 最深部でないのに、アクティブなノードが0になったときの枝切り
			if (i != N - 1 && nowTop == 0) {
				System.out.println(-1);
				System.exit(0);
			}
			// アクティブなノードが0未満の枝切り
			if (nowTop < 0) {
				System.out.println(-1);
				System.exit(0);
			}

			if (div > 0) {
				// 可能な限り分ける
				long nowDiv = Math.min(nowTop, div);
				div -= nowDiv;
				nowTop += nowDiv;
				numOfTop += nowTop;
				// 分岐できない頂点を除去
				nowTop -= input[i + 1];
			} else {
				// 分けない
				numOfTop += nowTop;
				// 分岐できない頂点を除去
				nowTop -= input[i + 1];
			}
		}
		if (nowTop == 0) {
			System.out.println(numOfTop);
		} else {
			System.out.println(-1);
		}
	}

}
