You are given a tree (a graph with n vertices and n - 1 edges in which it's possible to reach any vertex from any other vertex using only its edges).

A vertex can be destroyed if this vertex has even degree. If you destroy a vertex, all edges connected to it are also deleted.

Destroy all vertices in the given tree or determine that it is impossible.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — number of vertices in a tree.

The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ n). If pi ≠ 0 there is an edge between vertices i and pi. It is guaranteed that the given graph is a tree.

Output

If it's possible to destroy all vertices, print "YES" (without quotes), otherwise print "NO" (without quotes).

If it's possible to destroy all vertices, in the next n lines print the indices of the vertices in order you destroy them. If there are multiple correct answers, print any.

Examples

Input

5
0 1 2 1 2


Output

YES
1
2
3
5
4


Input

4
0 1 2 3


Output

NO

Note

In the first example at first you have to remove the vertex with index 1 (after that, the edges (1, 2) and (1, 4) are removed), then the vertex with index 2 (and edges (2, 3) and (2, 5) are removed). After that there are no edges in the tree, so you can remove remaining vertices in any order.

<image>
#include <bits/stdc++.h>
using namespace std;
vector<int> adj[200000 + 20];
bool mitParentKante[200000 + 20];
bool ohneParentKante[200000 + 20];
void dfs(int i, int parent) {
  if (adj[i].size() == 1 && parent != -1) {
    ohneParentKante[i] = true;
    mitParentKante[i] = false;
    return;
  }
  bool geradeAnzahlOhne = true;
  for (int child : adj[i]) {
    if (child != parent) {
      dfs(child, i);
      if (!mitParentKante[child] && !ohneParentKante[child]) {
        cout << "NO" << endl;
        exit(0);
      }
      if (ohneParentKante[child]) {
        geradeAnzahlOhne = !geradeAnzahlOhne;
      }
      if (mitParentKante[child]) {
      }
    }
  }
  if (geradeAnzahlOhne) {
    ohneParentKante[i] = true;
    mitParentKante[i] = false;
  } else {
    ohneParentKante[i] = false;
    mitParentKante[i] = true;
  };
}
void deleteDFS(int i, int parent) {
  for (int child : adj[i]) {
    if (child != parent) {
      ;
      ;
      ;
      if (mitParentKante[child]) {
        deleteDFS(child, i);
      }
    }
  }
  cout << i + 1 << endl;
  for (int child : adj[i]) {
    if (child != parent) {
      if (ohneParentKante[child]) {
        deleteDFS(child, i);
      }
    }
  }
}
int32_t main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int v;
    cin >> v;
    v--;
    if (v != -1) {
      adj[v].push_back(i);
      adj[i].push_back(v);
    }
  }
  dfs(0, -1);
  if (!ohneParentKante[0]) {
    cout << "NO" << endl;
    exit(0);
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    ;
    ;
    ;
  }
  deleteDFS(0, -1);
}
