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;
int n;
void addedge(vector<int> adj[], int a, int b) {
  adj[a].push_back(b);
  adj[b].push_back(a);
}
deque<int> st;
void printst() {
  for (auto it = st.begin(); it != st.end(); ++it) {
    cout << *it << " ";
  }
  cout << "\n";
}
void dfs(vector<int> adj[], int a, vector<bool> &visited) {
  visited[a] = true;
  st.push_back(a);
  for (int i = 0; i < (int)adj[a].size(); ++i) {
    if (!visited[adj[a][i]]) {
      if (adj[a][i] == n) {
        st.push_back(n);
        printst();
      }
      dfs(adj, adj[a][i], visited);
    }
  }
  st.pop_back();
}
int main() {
  cin >> n;
  vector<int> adj[n + 1];
  for (int i = 2; i < n + 1; ++i) {
    int temp;
    cin >> temp;
    addedge(adj, temp, i);
  }
  vector<bool> visited(n + 1, false);
  dfs(adj, 1, visited);
}
