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 
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 17 13:44:48 2020

@author: pctab
"""
import sys
sys.setrecursionlimit(99999999)
from collections import defaultdict
graph=defaultdict(set)
n=int(input())
arr=list(map(int,input().split()))
for i in range(len(arr)):
    graph[arr[i]].add(i+2)
parent=[-1]*(n+1)

def bfs():
    from collections import deque
    q=deque()
    q.append(1)
    visited=set()
    
    while q:
        u=q.popleft()
        visited.add(u)
        for v in graph[u]:
            if v not in visited:
                parent[v]=u
                q.append(v)
bfs()
path=[n]
v=n
while v!=1:    
    v=parent[v]
    path.append(v)
path.reverse()
for x in path:
    print(x,end=' ')
    
