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;
const int N = 2 * 100 * 1000 + 10;
int n, startPoint;
int sz[N], out[N];
vector<int> child[N], in[N];
void input() {
  cin.tie(0);
  cin >> n;
  for (int i = 0, u; i < n; i++) {
    cin >> u;
    if (u != 0)
      child[u - 1].push_back(i);
    else
      startPoint = i;
  }
  return;
}
int DFS(int v) {
  int res = 1;
  for (auto u : child[v]) {
    int tmp = DFS(u);
    if (tmp % 2) {
      in[v].push_back(u);
      out[u]++;
    } else {
      in[u].push_back(v);
      out[v]++;
    }
    res += tmp;
  }
  return res;
}
void solve() {
  cout.tie(0);
  cout << "YES\n";
  set<pair<int, int> > st;
  for (int i = 0; i < n; i++) st.insert({out[i], i});
  for (int i = 0; i < n; i++) {
    pair<int, int> p = *(st.begin());
    st.erase(st.begin());
    cout << p.second + 1 << '\n';
    for (auto u : in[p.second]) {
      st.erase({out[u], u});
      st.insert({--out[u], u});
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  input();
  if (n % 2) {
    DFS(startPoint);
    solve();
  } else
    cout << "NO" << endl;
  return 0;
}
