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 nxi() {
  int a;
  cin >> a;
  return a;
}
vector<int> a;
void inc(int i, int v) {
  for (; i < a.size(); i |= i + 1) {
    a[i] = max(a[i], v);
  }
}
int get(int r) {
  int ans = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    ans = max(ans, a[r]);
  }
  return ans;
}
int get(int l, int r) { return get(r) - get(l - 1); }
int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); }
const int mod = 1000000009;
int main() {
  int n = nxi();
  vector<vector<int> > adj(n);
  for (int i = 0; i < n; ++i) {
    int p = nxi();
    if (p != 0) {
      adj[p - 1].push_back(i);
      adj[i].push_back(p - 1);
    }
  }
  vector<int> dyn(n, 0);
  vector<int> answer;
  vector<int> q;
  function<void(int, int)> rec = [&](int cur, int parent) {
    int children = 0;
    int pos = 0;
    for (int v : adj[cur]) {
      if (v != parent) {
        children++;
        rec(v, cur);
        pos += dyn[v];
        if (dyn[v] == 0) {
          answer.push_back(v);
        } else {
          q.push_back(v);
        }
      }
    }
    dyn[cur] = pos % 2 == 0;
  };
  rec(0, -1);
  q.push_back(0);
  if (dyn[0]) {
    cout << "YES\n";
    for (int v : answer) cout << (v + 1) << endl;
    reverse(q.begin(), q.end());
    for (int v : q) cout << (v + 1) << endl;
  } else {
    cout << "NO\n";
  }
  return 0;
}
