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<vector<long long>> g;
vector<int> color, parent;
void dfs(int u, int from) {
  color[u] = 1;
  parent[u] = from;
  for (long long i = 0; i < g[u].size(); ++i) {
    long long v = g[u][i];
    if (color[v] == 0) dfs(v, u);
  }
}
int main() {
  long long n;
  cin >> n;
  g.resize(n);
  color.resize(n, 0);
  parent.resize(n);
  for (long long i = 0; i < n - 1; ++i) {
    long long p;
    cin >> p;
    g[p - 1].push_back(i + 1);
  }
  long long start = 0, finish = n - 1;
  dfs(start, -1);
  vector<long long> ans;
  for (long long u = finish; u != start; u = parent[u]) ans.push_back(u);
  ans.push_back(start);
  reverse(ans.begin(), ans.end());
  long long i;
  for (i = 0; i < ans.size() - 1; ++i) cout << ans[i] + 1 << " ";
  cout << ans[i] + 1;
  return 0;
}
