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 dfs(vector<int> g[], int n, int at, vector<int>& res, vector<bool>& vis) {
  vis[at] = 1;
  res.push_back(at);
  if (at == n) {
    for (int x : res) {
      cout << x << " ";
    }
    return;
  }
  for (auto i = g[at].begin(); i != g[at].end(); i++) {
    if (!vis[*i]) {
      dfs(g, n, *i, res, vis);
    }
    res.erase(res.end() - 1, res.end());
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> g[n];
  vector<bool> vis(2000000);
  vector<int> res;
  fill(vis.begin(), vis.end(), 0);
  for (int i = 2; i <= n; i++) {
    int u;
    cin >> u;
    g[u].push_back(i);
  }
  dfs(g, n, 1, res, vis);
}
