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 comment(linker, "/STACK:102400000,102400000")
using namespace std;
const double eps = 1e-6;
inline long long pow(long long a, long long b, long long p) {
  long long rtn = 1;
  while (b) {
    if (b & 1) rtn = rtn * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return rtn;
}
long long inv(long long a, long long p) { return pow(a, p - 2, p); }
long long gcd(long long a, long long b) {
  long long k;
  while (b != 0) {
    k = b;
    b = a % b;
    a = k;
  }
  return a;
}
void makedata() {
  freopen("input.txt", "w", stdout);
  srand((unsigned)time(NULL));
  fclose(stdout);
}
int fa[220000], p[220000];
int main() {
  std::ios::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  for (int i = (2); i <= (n); i++) {
    cin >> fa[i];
  }
  int m = 0, x = n;
  while (x != 1) {
    p[m++] = x;
    x = fa[x];
  }
  cout << (1) << ' ';
  for (int i = (m - 1); i >= (0); i--) cout << (p[i]) << ' ';
  return 0;
}
