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;
const long long int mod = 212072634227239451;
vector<long long int> ans;
bool prime(long long int s) {
  if (s <= 1) return false;
  if (s <= 3) return true;
  if (s % 2 == 0 || s % 3 == 0) return false;
  for (long long int i = 5; i * i <= s; i = i + 6)
    if (s % i == 0 || s % (i + 2) == 0) return false;
  return true;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool square(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
const long long int sz = 1e6;
bool vis[sz] = {false};
long long int par[sz];
void dfs(vector<long long int> adj[], long long int u, long long int n) {
  vis[u] = true;
  if (u == n) return;
  for (long long int chd : adj[u]) {
    if (!vis[chd]) {
      par[chd] = u;
      dfs(adj, chd, n);
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  long long int u;
  vector<long long int> adj[n + 1];
  for (long long int i = 2; i <= n; ++i)
    cin >> u, adj[i].emplace_back(u), adj[u].emplace_back(i);
  par[1] = 1;
  dfs(adj, 1, n);
  long long int d = n;
  vector<long long int> ans;
  while (n != 1) {
    ans.emplace_back(n);
    n = par[n];
  }
  ans.emplace_back(1);
  reverse(ans.begin(), ans.end());
  for (long long int num : ans) cout << num << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
