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;
vector<set<int> > adj(222222);
vector<int> ans;
void doit(int x, int p) {
  if (p != -1) {
    adj[p].erase(adj[p].find(x));
    adj[x].erase(adj[x].find(p));
  }
  ans.push_back(x);
  vector<int> edg;
  for (__typeof(adj[x].begin()) it = adj[x].begin(); it != adj[x].end(); it++)
    edg.push_back(*it);
  for (__typeof(edg.begin()) it = edg.begin(); it != edg.end(); it++)
    doit(*it, x);
}
void dfs(int x, int p) {
  int ec = 0;
  vector<int> edg;
  for (__typeof(adj[x].begin()) it = adj[x].begin(); it != adj[x].end(); it++) {
    if (*it == p) continue;
    edg.push_back(*it);
  }
  for (__typeof(edg.begin()) it = edg.begin(); it != edg.end(); it++)
    dfs(*it, x);
  ec += adj[x].size();
  if (ec % 2 == 0) doit(x, p);
}
int main() {
  cin >> n;
  int x;
  for (int i = 1; i <= n; i++) {
    cin >> (x);
    if (x == 0) continue;
    adj[i].insert(x);
    adj[x].insert(i);
  }
  dfs(1, -1);
  if (adj[1].size() == 0) {
    cout << "YES" << '\n';
    for (__typeof(ans.begin()) it = ans.begin(); it != ans.end(); it++)
      cout << *it << '\n';
  } else
    cout << "NO" << '\n';
}
