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;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int n;
vector<int> v[maxn];
vector<int> ans;
bool vis[maxn];
bool flag = 1;
void dfs(int k) {
  vis[k] = 1;
  if (k == n) {
    ans.push_back(k);
    flag = 0;
    return;
  }
  int siz = v[k].size();
  for (int i = 0; i < siz; i++) {
    if (!vis[v[k][i]] && flag) {
      dfs(v[k][i]);
    }
  }
  if (!flag) ans.push_back(k);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int temp;
    scanf("%d", &temp);
    v[temp].push_back(i);
  }
  dfs(1);
  int siz = ans.size();
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < siz; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
