Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).

There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.

Input

The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.

Output

Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.

Examples

Input

8
1 1 2 2 3 2 5


Output

1 2 5 8 

Input

6
1 2 3 4 5


Output

1 2 3 4 5 6 

Input

7
1 1 2 3 4 3


Output

1 3 7 
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class source4 {

	static class Router {
		public int parentPos;
		public Router parent;
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();

		Router[] rs = new Router[n + 1];
		rs[1] = new Router();

		for (int i = 2; i <= n; i++) {
			Router r = new Router();
			r.parentPos = in.nextInt();
			rs[i] = r;
		}

		for (int i = 2; i <= n; i++) {
			rs[i].parent = rs[rs[i].parentPos];
		}

		ArrayList<Integer> rList = new ArrayList<Integer>();
		Router rout = rs[n];
		while (rout.parent != null) {
			rList.add(rout.parentPos);
			rout = rout.parent;
		}

		Collections.reverse(rList);
		rList.add(n);

		for (int i = 0, z = rList.size(); i < z; i++) {
			if (i == z - 1) {
				System.out.print(rList.get(i));
				break;
			}
			System.out.print(rList.get(i) + " ");
		}
		in.close();
	}
}
