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 
from sys import stdin
getc = lambda: stdin.read(1)
def alph():
    a = list(range(65, 91))
    b = list(range(97, 123))
    return set(([chr(i) for i in a + b]))
def cin(T = "int"):
    d = {"int" : {'1', '2', '3', '4', '5', '6', '7', '8', '9' ,'0'}, "str" : alph()}
    a = getc()
    res = ''
    while a in d[T]:
        res += a
        a = getc()
    if T == 'int':
        return int(res)
    return res
def printv(a, sep = '\n'):
    print(sep.join(map(str, a)))
def bruh():
    n = cin("int")
    a = [cin("int")for i in range(n - 1)]
    ans = []
    cur = n
    while(cur != 1):
        ans += [cur]
        cur = a[cur - 2]
    ans += [1]
    printv(ans[:: -1], ' ')
bruh()
