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>
const long long maxn = 2 * 100001;
using namespace std;
vector<long long> edge[maxn];
bool vis[maxn];
void dfs(long long x, set<long long> &ans, long long n, bool &flag) {
  vis[x] = true;
  if (x == n) {
    flag = 1;
    ans.insert(x);
    return;
  }
  for (auto &i : edge[x]) {
    if (!vis[i]) {
      dfs(i, ans, n, flag);
      if (flag) {
        ans.insert(x);
        return;
      }
    }
  }
}
void solve() {
  long long n, x;
  memset(vis, false, sizeof(vis));
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    cin >> x;
    edge[x].push_back(i);
  }
  set<long long> ans;
  bool flag = false;
  dfs(1, ans, n, flag);
  for (auto &x : ans) {
    cout << x << " ";
  }
  cout << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
