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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T maxs(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T mins(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long solve() {
  long long n;
  cin >> n;
  vector<long long> g[n + 1];
  long long r;
  for (long long i = 1; i < n + 1; i++) {
    long long p;
    cin >> p;
    if (p) g[p].push_back(i);
    if (!p) r = i;
  }
  vector<long long> dp(n + 1), h(n + 1);
  vector<long long> vec;
  function<void(long long)> dfs = [&](long long u) {
    long long d = 0;
    if (u != r) d++;
    for (auto i : g[u]) {
      h[i] = h[u] + 1;
      dfs(i);
      if (dp[i] == 0) d++;
    }
    if (d % 2 == 0) {
      vec.push_back(u);
      dp[u] = 1;
    } else
      dp[u] = 0;
  };
  dfs(r);
  if (dp[r]) {
    cout << "YES"
         << "\n";
    vector<pair<long long, long long> > p;
    for (long long i = 1; i < n + 1; i++) {
      if (dp[i] == 0) p.push_back({h[i], i});
    }
    sort(p.begin(), p.end());
    for (auto i : vec) cout << i << "\n";
    for (auto i : p) cout << i.second << "\n";
  } else {
    cout << "NO"
         << "\n";
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
