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>
#pragma GCC optimize("Ofast")
using namespace std;
void fastaf(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
vector<long long int> children[1000000];
vector<long long int> path;
long long int N;
long long int flag = 0;
void dfs(long long int current, long long int parent, long long int &flag) {
  if (current == N - 1) {
    flag = 1;
    path.push_back(current);
  }
  for (auto x : children[current]) {
    dfs(x, current, flag);
    if (flag == 1) {
      path.push_back(current);
      return;
    }
  }
}
void solve(void) {
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    long long int x;
    cin >> x;
    x--;
    children[x].push_back(i + 1);
  }
  flag = 0;
  dfs(0, -1, flag);
  reverse(path.begin(), path.end());
  for (int i = 0; i < path.size(); i++) cout << path[i] + 1 << " ";
}
int main() {
  fastaf();
  solve();
  cout << endl;
  return 0;
}
