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;
void printpath(vector<int> v) {
  int i = 0;
  for (i = 0; i < v.size() - 1; i++) {
    cout << v[v.size() - i - 1] << " ";
  }
  cout << v[v.size() - i - 1];
}
void dfs(int s, int d, vector<int> adj[], vector<int> &vis, vector<int> &v) {
  v.push_back(d);
  if (s == d) {
    printpath(v);
    return;
  }
  int flag = 0;
  vis[d] = 1;
  for (int i = 0; i < adj[d].size(); i++) {
    if (vis[adj[d][i]] == 0) {
      dfs(s, adj[d][i], adj, vis, v);
      flag = 1;
    }
  }
  if (flag == 0) v.pop_back();
}
int main() {
  int n;
  cin >> n;
  vector<int> adj[n + 1];
  int temp;
  vector<int> v;
  for (int i = 2; i <= n; i++) {
    cin >> temp;
    adj[temp].push_back(i);
    adj[i].push_back(temp);
  }
  vector<int> vis(n + 1, 0);
  dfs(1, n, adj, vis, v);
}
