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<vector<bool> > dp;
vector<bool> en, used;
vector<int> ans, ans1;
int st;
void dfs(int v) {
  int c = 0;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    dfs(to);
    if (!used[to]) {
      c++;
    }
  }
  if (c % 2 == 1) {
    if (v == st) {
      cout << "NO";
      exit(0);
    }
    used[v] = 1;
    ans.push_back(v);
  } else {
    ans1.push_back(v);
  }
}
signed main() {
  int n;
  cin >> n;
  used.resize(n);
  en.resize(n);
  dp.resize(n, vector<bool>(2));
  g.resize(n);
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    p--;
    if (p == -1) {
      st = i;
    } else {
      g[p].push_back(i);
    }
  }
  dfs(st);
  cout << "YES\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << "\n";
  }
  for (int i = ans1.size() - 1; i >= 0; i--) {
    cout << ans1[i] + 1 << "\n";
  }
  return 0;
}