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;
string dfs(vector<int> adj[], int n, int i, vector<int> &visited) {
  queue<pair<int, string>> q;
  string ans = "";
  string temp = "1";
  q.push({i, temp});
  while (q.size() > 0) {
    int node = q.front().first;
    temp = q.front().second;
    visited[node] = 1;
    q.pop();
    if (node == n - 1) {
      ans = temp;
      break;
    }
    for (auto j : adj[node]) {
      if (!visited[j]) {
        q.push({j, temp + " " + to_string(j + 1)});
      }
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    int n;
    cin >> n;
    map<int, int> connection;
    int x;
    for (int i = 2; i <= n; ++i) {
      cin >> x;
      connection[i] = x;
    }
    int node = n;
    vector<int> ans;
    ans.push_back(node);
    while (node != 1) {
      node = connection[node];
      ans.push_back(node);
    }
    reverse(ans.begin(), ans.end());
    for (auto i : ans) cout << i << " ";
  }
  return 0;
}
