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;
using ll = long long;
template <typename... t>
using V = vector<t...>;
template <typename... t>
using outit = ostream_iterator<t...>;
void print(ostream& os) { os << '\n'; }
template <typename t, typename... v>
void print(ostream& os, const t& a, v&&... b) {
  os << a << ' ';
  print(os, b...);
}
int n;
V<int> odp;
V<int> ilo, poz;
V<V<int> > gr;
V<V<bool> > dp;
bool dfs1(int v, int oj) {
  if (static_cast<int>((gr[v]).size()) == 1 and v != oj) return dp[1][v] = true;
  for (int s : gr[v])
    if (s != oj)
      if (not dfs1(s, v)) return false;
  for (int s : gr[v])
    if (s != oj) {
      if (dp[0][s] and dp[1][s])
        ilo[v] = s;
      else if (dp[1][s])
        ++poz[v];
    }
  dp[0][v] = ((poz[v] & 1) == 1) or ilo[v] != -1;
  dp[1][v] = ((poz[v] & 1) == 0) or ilo[v] != -1;
  if ((not dp[0][v] and not dp[1][v]) or (v == oj and not dp[1][v]))
    return false;
  return true;
}
void dfs2(int v, int oj, bool fl = false) {
  bool zm = false;
  if (fl == false and (poz[v] & 1) == 1) zm = true;
  if (fl == true and (poz[v] & 1) == 0) zm = true;
  for (int s : gr[v])
    if (s != oj) {
      if (zm and s == ilo[v]) continue;
      if (dp[0][s]) dfs2(s, v, false);
    }
  odp.push_back(v);
  for (int s : gr[v])
    if (s != oj) {
      if (zm and s == ilo[v]) {
        dfs2(s, v, true);
        continue;
      }
      if (dp[0][s] == false and dp[1][s] == true) dfs2(s, v, true);
    }
}
int main(int argc, char** argv) {
  (void)(argc);
  (void)(argv);
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  gr.resize(n + 1);
  ilo.resize(n + 1, -1);
  poz.resize(n + 1);
  dp.resize(2, V<bool>(n + 1));
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p;
    if (p != 0) {
      gr[i].push_back(p);
      gr[p].push_back(i);
    }
  }
  if (not dfs1(1, 1)) return print(cout, "NO"), 0;
  ;
  dfs2(1, 1);
  print(cout, "YES");
  copy(begin(odp), end(odp), outit<int>(cout, "\n"));
}