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;
int p[200010], d[200010];
vector<int> e[200010], ord;
void dfs(int u) {
  for (int v : e[u]) dfs(v);
  ord.push_back(u);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  memset(d, 0, sizeof d);
  int root = 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    d[p[i]]++;
    if (p[i])
      d[i]++;
    else
      root = i;
    e[p[i]].push_back(i);
  }
  dfs(root);
  vector<int> e, ans;
  for (auto i : ord) {
    if (d[i] % 2) {
      if (i == root) {
        cout << "NO\n";
        return 0;
      }
      e.push_back(i);
    } else {
      ans.push_back(i);
      d[p[i]]--;
    }
  }
  reverse(e.begin(), e.end());
  ans.insert(ans.end(), e.begin(), e.end());
  cout << "YES\n";
  for (int x : ans) {
    cout << x << "\n";
  }
  return 0;
}
