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 int N = 2e5 + 5;
vector<int> adj[N];
int n;
bool dfs(int node, vector<int>& ans, int parent = -1) {
  ans.push_back(node);
  if (node == n) return true;
  for (auto child : adj[node])
    if (child != parent)
      if (dfs(child, ans, node)) return true;
  ans.pop_back();
  return false;
}
void solve() {
  cin >> n;
  vector<int> ar(n - 1);
  for (int i = 0; i + 1 < n; i++) {
    cin >> ar[i];
    adj[i + 2].push_back(ar[i]);
    adj[ar[i]].push_back(i + 2);
  }
  vector<int> ans;
  dfs(1, ans);
  for (auto x : ans) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
