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;
vector<int> adj[200003];
int parent[200003];
bool vis[200003];
int n;
void bfs(int i) {
  queue<int> q;
  q.push(i);
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    for (auto v : adj[p]) {
      if (!vis[v]) {
        parent[v] = p;
        vis[v] = true;
        q.push(v);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x;
    cin >> x;
    adj[x].push_back(i + 2);
    adj[i + 2].push_back(x);
  }
  bfs(1);
  vector<int> ans;
  parent[1] = 0;
  for (int i = n; i >= 1; i = parent[i]) {
    ans.push_back(i);
  }
  for (int i = ans.size() - 1; i >= 0; --i) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
