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 namespace chrono;
vector<int> s;
vector<int> ar[200001];
void dfs(int node, int n) {
  s.push_back(node);
  if (node == n) {
    for (int i = 0; i < s.size() - 1; i++) {
      cout << s[i] << " ";
    }
    cout << n;
  }
  for (int child : ar[node]) {
    dfs(child, n);
  }
  s.pop_back();
}
void solve() {
  int n, p;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> p;
    ar[p].push_back(i);
  }
  dfs(1, n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start1 = high_resolution_clock::now();
  solve();
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
}
