Problem: 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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void get_destructibility(int v, int par, vector<set<int> >& edges,
                         vector<bool>& is_destructible) {
  is_destructible[v] = true;
  for (int w : edges[v]) {
    if (w != par) {
      get_destructibility(w, v, edges, is_destructible);
      if (is_destructible[w]) {
        is_destructible[v] = !is_destructible[v];
      }
    }
  }
}
void get_destruction_order(int v, int par, vector<set<int> >& edges,
                           vector<bool>& is_destructible,
                           list<int>& destruction_order) {
  for (int w : edges[v]) {
    if (w != par && !is_destructible[w]) {
      get_destruction_order(w, v, edges, is_destructible, destruction_order);
    }
  }
  destruction_order.push_back(v);
  for (int w : edges[v]) {
    if (w != par && is_destructible[w]) {
      get_destruction_order(w, v, edges, is_destructible, destruction_order);
    }
  }
}
int main() {
  int n;
  cin >> n;
  vector<set<int> > edges(n);
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    if (p) {
      edges[i].insert(p - 1);
      edges[p - 1].insert(i);
    }
  }
  vector<bool> is_destructible(n);
  get_destructibility(0, -1, edges, is_destructible);
  if (is_destructible[0]) {
    cout << "YES\n";
    list<int> destruction_order;
    get_destruction_order(0, -1, edges, is_destructible, destruction_order);
    for (int v : destruction_order) {
      cout << v + 1 << '\n';
    }
  } else {
    cout << "NO\n";
  }
}