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;
int n, x, root, dis[200005], vis[200005];
set<int> adj[200005];
set<pair<int, int> > st;
vector<int> print;
void bfs(void) {
  queue<int> q;
  q.push(root);
  while (q.size()) {
    int u = q.front();
    q.pop();
    set<int>::iterator tt;
    for (tt = adj[u].begin(); tt != adj[u].end(); tt++) {
      int xx = *tt;
      if (xx == root || dis[xx]) continue;
      dis[xx] = dis[u] + 1;
      q.push(xx);
    }
  }
}
void _delete(int xx) {
  print.push_back(xx);
  set<int>::iterator tt;
  for (tt = adj[xx].begin(); tt != adj[xx].end(); tt++) {
    int z = *tt;
    adj[z].erase(xx);
    if (vis[z] && adj[z].size() % 2 == 0) _delete(z);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x == 0)
      root = i;
    else {
      adj[i].insert(x);
      adj[x].insert(i);
    }
  }
  bfs();
  set<pair<int, int> >::iterator it;
  for (int i = 1; i <= n; i++) st.insert(make_pair(-dis[i], i));
  for (it = st.begin(); it != st.end(); it++) {
    int xx = it->second;
    vis[xx] = 1;
    if ((adj[xx].size() * 1LL) % 2) continue;
    _delete(xx);
  }
  if (print.size() != n) cout << "NO" << endl, exit(0);
  cout << "YES" << endl;
  for (int i = 0; i < print.size(); i++) cout << print[i] << "\n";
  return 0;
}
