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;
long long mod = 1e9 + 7;
const int N = 2e5 + 5;
vector<vector<int>> g;
bool vis[N];
int n;
bool dfs(int v, vector<int>& path) {
  if (vis[v]) return false;
  vis[v] = true;
  bool f = false;
  for (auto u : g[v]) {
    f = f || dfs(u, path);
  }
  if (v == n) f = true;
  if (f) path.push_back(v);
  return f;
}
void solve() {
  int x;
  cin >> n;
  g.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    cin >> x;
    g[x].push_back(i + 2);
  }
  vector<int> path;
  dfs(1, path);
  for (auto rit = path.rbegin(); rit != path.rend(); rit++) {
    cout << (*rit) << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
