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;
vector<int> v, v1;
vector<int> a[200005], b[200004];
int vis[200004], vis1[200004];
int dfs(int i) {
  vis[i] = 1;
  int cnt = 0;
  for (int j = 0; j < a[i].size(); j++) {
    if (vis[a[i][j]] == 0) {
      int z = dfs(a[i][j]);
      cnt += z;
      if (z == 1)
        b[i].push_back(a[i][j]);
      else
        b[a[i][j]].push_back(i);
    }
  }
  v.push_back(i);
  return 1 - (cnt % 2);
}
void dfs2(int i) {
  vis1[i] = 1;
  for (int j = 0; j < b[i].size(); j++) {
    if (vis1[b[i][j]] == 0) {
      dfs2(b[i][j]);
    }
  }
  v1.push_back(i);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p != 0) {
      a[p].push_back(i);
      a[i].push_back(p);
    }
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      v.clear();
      v1.clear();
      int w = dfs(i);
      if (w == 0) {
        cout << "NO" << '\n';
        return 0;
      } else {
        for (int j = 0; j < v.size(); j++) {
          if (vis1[v[j]] == 0) {
            dfs2(v[j]);
          }
        }
        for (int j = 0; j < v1.size(); j++) ans.push_back(v1[j]);
      }
    }
  }
  cout << "YES" << '\n';
  for (int i = ans.size() - 1; i >= 0; i--) cout << ans[i] << '\n';
}
