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>
#pragma GCC optimise("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,abm,mmx,popcnt,avx,avx2,tune=native")
#pragma comment(linker, "/STACK:128777216")
using namespace std;
using namespace std::chrono;
int const N = 2e5;
vector<int> gr[N];
vector<bool> used(N);
int n;
vector<int> ans;
void dfs(int v) {
  used[v] = true;
  ans.push_back(v);
  if (v == n - 1) {
    for (auto elem : ans) {
      cout << elem + 1 << ' ';
    }
    exit(0);
  }
  for (auto to : gr[v]) {
    if (!used[to]) {
      dfs(to);
    }
  }
  ans.pop_back();
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x;
    cin >> x;
    gr[i].push_back(x - 1);
    gr[x - 1].push_back(i);
  }
  dfs(0);
  return 0;
}
