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 
#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-13;
const long double PI = acos(-1);
const int INF = (int)1e9;
const long long INFF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MXN = (int)1e6 + 7;
int prt[MXN];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 2; i < n + 1; i++) scanf("%d", prt + i);
  int cp = n;
  vector<int> ans;
  while (cp) {
    ans.push_back(cp);
    cp = prt[cp];
  }
  reverse(begin(ans), end(ans));
  for (int i = 0; i < ((int)(ans).size()); i++)
    printf("%d%c", ans[i], " \n"[i == ((int)(ans).size()) - 1]);
  return 0;
}
