Problem: 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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > adj[200005];
int n;
int dfs1(int i, int p) {
  int sum = 1;
  for (auto& j : adj[i]) {
    if (j.first != p) {
      j.second = dfs1(j.first, i);
      sum += j.second;
    }
  }
  return sum;
}
int dfs2(int i, int p) {
  for (auto j : adj[i]) {
    if (j.first != p && !(j.second & 1)) {
      dfs2(j.first, i);
    }
  }
  cout << i << '\n';
  for (auto j : adj[i]) {
    if (j.first != p && (j.second & 1)) {
      dfs2(j.first, i);
    }
  }
}
int main() {
  cin >> n;
  if (!(n & 1)) cout << "NO\n", exit(0);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    int q;
    cin >> q;
    if (q) {
      adj[i].push_back(make_pair(q, 0));
      adj[q].push_back(make_pair(i, 0));
    }
  }
  dfs1(1, 0);
  dfs2(1, 0);
}