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;
template <typename XTYPE>
inline void _var_(const XTYPE& a) {}
template <typename XTYPE>
inline void _vector_(const XTYPE& a) {}
template <typename XTYPE>
inline void _matrix_(const XTYPE& a) {}
inline void fast_input() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
}
const long long LINF = LLONG_MAX;
const int INF = 1000000000;
const double PI = atan(1.0) * 4;
const int MOD = 1e9 + 7;
int n;
vector<int> p;
int main() {
  cin >> n;
  p.resize(n + 1);
  for (int i = 0; i < int(n - 1); i++) {
    int buf;
    cin >> buf;
    buf--;
    p[i + 1] = buf;
  }
  vector<int> ans;
  ans.reserve(n);
  int cur = n - 1;
  ans.push_back(cur);
  while (p[cur] != 0) {
    ans.push_back(p[cur]);
    cur = p[cur];
  }
  ans.push_back(0);
  reverse(ans.begin(), ans.end());
  for (auto t : ans) {
    cout << t + 1 << " ";
  }
  return 0;
}
