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;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  int l = v.size();
  for (int i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int N = 200200;
int done[N];
vector<int> g[N];
vector<int> order;
void purge(int x, int prev) {
  if (done[x]) return;
  order.push_back(x);
  done[x] = 1;
  for (int y : g[x]) {
    if (y == prev) continue;
    purge(y, x);
  }
}
bool dfs(int x, int prev) {
  bool leaf = true;
  int par = 0;
  for (int y : g[x]) {
    if (y == prev) continue;
    leaf = false;
    if (!dfs(y, x)) par ^= 1;
  }
  if (leaf) return false;
  if (prev != -1) {
    if (par) {
      purge(x, prev);
      return true;
    }
    return false;
  } else {
    if (par) return false;
    purge(x, prev);
    return true;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  if (n == 1) {
    cout << "YES" << endl;
    cout << 1 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x) {
      g[x].push_back(i);
      g[i].push_back(x);
    }
  }
  bool res = dfs(1, -1);
  if (!res) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (int x : order) {
      cout << x << '\n';
    }
  }
  return 0;
}