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;
const int mxN = 1e5 + 1, MIN = -1e9, MAX = 1e9;
void yes() {
  cout << "YES\n";
  return;
}
void no() {
  cout << "NO\n";
  return;
}
void out(long long int n) {
  cout << n << endl;
  return;
}
void outs(string str) {
  cout << str << endl;
  return;
}
void init() {}
void solve(int case_no) {
  int n;
  cin >> n;
  int a[n + 1];
  a[1] = 0;
  for (int i = 2; i <= n; i++) cin >> a[i];
  int ptr = n;
  stack<int> s;
  while (ptr != 0) {
    s.push(ptr);
    ptr = a[ptr];
  }
  while (!s.empty()) {
    cout << s.top() << ' ';
    s.pop();
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(NULL));
  init();
  int t = 1;
  for (int i = 1; i <= t; i++) solve(i);
}
