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;
priority_queue<long long, std::vector<long long>, std::greater<long long> > pqs;
queue<long long> qq;
stack<long long> sta;
set<long long> st;
set<long long>::iterator P;
pair<long long, long long> pp;
vector<long long> v[100000];
map<long long, long long> mp, pm;
long long i, j, n, m, t, k, a, b, c, x, y, mx, mn;
string s, s1;
char z;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int T = 1;
  while (T--) {
    i = j = n = m = t = k = a = b = c = x = y = mx = mn = 0;
    cin >> n;
    long long A[n + 5], B[n + 5];
    for (i = 2; i <= n; i++) cin >> A[i];
    for (i = n; i > 1; i = A[i]) B[j] = i, j++;
    cout << 1;
    for (i = j - 1; i >= 0; i--) cout << " " << B[i];
    cout << endl;
  }
  return 0;
}
