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> g[212345];
bitset<212345> destroyed;
int vcant[212345];
int solve1(int nodo, int cant) {
  for (auto it = g[nodo].begin(); it != g[nodo].end(); it++)
    cant += solve1(*it, 1);
  if (cant % 2) {
    vcant[nodo] = cant;
    return 1;
  }
  destroyed.set(nodo);
  for (auto it = g[nodo].begin(); it != g[nodo].end(); it++) vcant[*it]--;
  cout << nodo + 1 << '\n';
  return 0;
}
void solve2(int nodo) {
  bool aux = !destroyed[nodo] && vcant[nodo] % 2 == 0;
  if (aux) {
    cout << nodo + 1 << '\n';
    destroyed.set(nodo);
  }
  for (auto it = g[nodo].begin(); it != g[nodo].end(); it++) {
    if (aux) vcant[*it]--;
    solve2(*it);
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, p, root;
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = (0); i < (n); i++) {
    cin >> p;
    if (p == 0)
      root = i;
    else {
      g[p - 1].push_back(i);
    }
  }
  solve1(root, 0);
  solve2(root);
  for (int i = (0); i < (n); i++)
    if (!destroyed[i]) cout << i + 1 << '\n';
  return 0;
}
