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.AbstractQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;

public class Main {

    private static boolean used[];

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

        int n = in.nextInt();
        Stack<Integer> stack = new Stack<>();
        stack.push(n);

        int p[] = new int[n-1];

        for (int i = 0; i < p.length; i++) {
            p[i] = in.nextInt();
        }

        int i = 0;
        int m = n-2;

        while (i != 1) {
            i = p[m];
            m = i - 2;
            stack.push(i);
        }

        /*for (int i = 0; true; i++) {
            if (stack.peek() == p[i]) {
                stack.push(i + 2);
                stackPushPos.push(i);
            }
            if (stack.peek() == n)
                break;
            if (i+1 == p.length) {
                stack.pop();
                i = stackPushPos.peek();
                stackPushPos.pop();
            }
        }*/

        while (!stack.isEmpty()) {
            System.out.println(stack.pop());
        }

    }

}
