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<vector<int>> g;
vector<int> ans;
vector<int> kek;
void destroy(int v, int p) {
  if (kek[v] % 2 == 0) {
    kek[v] = -99999;
    ans.push_back(v);
    for (auto i : g[v]) {
      --kek[i];
      if (i != p) {
        destroy(i, v);
      }
    }
  }
}
void dfs(int v, int p) {
  for (auto i : g[v]) {
    if (p != i) {
      dfs(i, v);
    }
  }
  destroy(v, p);
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  kek.resize(n);
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    --a;
    if (a != -1) {
      g[i].push_back(a);
      g[a].push_back(i);
      kek[i]++;
      kek[a]++;
    }
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  dfs(0, -1);
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] + 1 << endl;
  }
}