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;
const long long MOD = 1e9 + 7;
const long long SZ = 107;
long long pow(long long x, long long y) {
  long long r = 1LL;
  while (y) {
    if (y & 1) r = (r * x) % MOD;
    y >>= 1;
    x = (x * x) % MOD;
  }
  return r;
}
long long add(long long x, long long y) {
  long long r = x + y;
  return r % MOD;
}
long long sub(long long x, long long y) {
  long long r = x - y;
  r = r % MOD;
  r = r + MOD;
  return r % MOD;
}
long long mul(long long x, long long y) {
  long long r = x * y;
  r %= MOD;
  r += MOD;
  return r % MOD;
}
long long inv(long long x, long long y) { return pow(x, y - 2); }
struct matrix {
  long long mat[SZ][SZ];
  void reset() { memset(mat, 0, sizeof(mat)); }
  void identical() {
    reset();
    for (int i = 0; i < SZ; i++) mat[i][i] = 1;
  }
  matrix operator+(const matrix &x) const {
    matrix r;
    for (int i = 0; i < SZ; i++)
      for (int j = 0; j < SZ; j++) r.mat[i][j] = add(mat[i][j], x.mat[i][j]);
    return r;
  }
  matrix operator*(const matrix &x) const {
    matrix r;
    for (int i = 0; i < SZ; i++)
      for (int j = 0; j < SZ; j++) {
        r.mat[i][j] = 0;
        for (int k = 0; k < SZ; k++)
          r.mat[i][j] = add(r.mat[i][j], mul(mat[i][k], x.mat[k][j]));
      }
    return r;
  }
};
matrix mpow(matrix x, int y) {
  matrix r;
  r.identical();
  while (y) {
    if (y & 1) r = r * x;
    x = x * x;
    y >>= 1;
  }
  return r;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    int p[n];
    for (int i = 2; i <= n; i++) {
      cin >> p[i];
    }
    int x = n;
    vector<int> ans;
    ans.push_back(x);
    while (x != 1) {
      x = p[x];
      ans.push_back(x);
    }
    reverse(ans.begin(), ans.end());
    for (auto i : ans) cout << i << " ";
  }
  return 0;
}
