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;
using i64 = unsigned long long;
using ll = long long;
using ld = long double;
using D = double;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
template <typename T>
T abs(T a) {
  return a > 0 ? a : -a;
}
const int inf = 2e9;
const ld st = 0.000001;
const ld pi = acos((ld)-1);
const int K = 1000;
const int NMAX = 1000;
int main() {
  int n;
  cin >> n;
  vector<vector<int> > e;
  e.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int x;
    cin >> x;
    e[i + 2].push_back(x);
    e[x].push_back(i + 2);
  }
  vector<int> d(n + 1, -1);
  vector<int> p(n + 1, 0);
  d[1] = 0;
  queue<int> q;
  q.push(1);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < e[v].size(); i++) {
      int to = e[v][i];
      if (d[to] == -1) {
        d[to] = d[v] + 1;
        q.push(to);
        p[to] = v;
      }
    }
  }
  stack<int> ans;
  ans.push(n);
  p[1] = -1;
  int s = p[n];
  while (s != -1) {
    ans.push(s);
    s = p[s];
  }
  while (!ans.empty()) {
    cout << ans.top() << " ";
    ans.pop();
  }
  return 0;
}
