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;
inline long long add(long long a, long long b, long long p = 1000000007) {
  long long c = a + b;
  if (c >= p) c -= p;
  return c;
}
inline long long sub(long long a, long long b, long long p = 1000000007) {
  long long c = a - b;
  if (c < 0) c += p;
  return c;
}
inline long long mul(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
vector<int> g[200005];
int t = 0;
int tim[500005];
void dfs(int u, int v = -1) {
  tim[t++] = u;
  for (auto x : g[u]) {
    if (x == v) continue;
    dfs(x, u);
  }
  tim[t++] = u;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  int a;
  cin >> n;
  int p[n + 1];
  for (int i = 2; i <= n; i++) {
    cin >> a;
    g[i].push_back(a);
    g[a].push_back(i);
  }
  dfs(1);
  map<int, int> mp;
  for (int i = 0; i <= 2 * n - 1; i++) {
    mp[tim[i]]++;
    if (tim[i] == n) break;
  }
  for (auto it = mp.begin(); it != mp.end(); it++) {
    if (it->second == 1) {
      cout << it->first << " ";
    }
  }
  return 0;
}
