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 p[200001];
vector<int> adj[200001];
int deg[200001];
int pt = 0;
int order[200001];
int ret[200001];
void dfs(int cur) {
  for (auto e : adj[cur]) dfs(e);
  order[++pt] = cur;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  if (n & 1)
    cout << "YES\n";
  else {
    cout << "NO\n";
    return 0;
  }
  int root = 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    adj[p[i]].push_back(i);
    if (p[i] == 0)
      root = i;
    else {
      deg[p[i]]++;
      deg[i]++;
    }
  }
  dfs(root);
  vector<int> ret;
  vector<int> ret2;
  for (int i = 1; i <= n; i++) {
    int e = order[i];
    if (deg[e] % 2 == 0) {
      ret.push_back(e);
      deg[p[e]]--;
    } else
      ret2.push_back(e);
  }
  ret.insert(ret.end(), ret2.rbegin(), ret2.rend());
  for (int i = 0; i < n; i++) cout << ret[i] << '\n';
  return 0;
}
