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;
const ll MOD = 1e9 + 7;
const int N = 2 * (int)1e5;
vector<int> conns[N];
int siz[N];
int dfs(int i, int p) {
  siz[i] = 1;
  for (auto t : conns[i]) {
    if (t != p) siz[i] += dfs(t, i);
  }
  return siz[i];
}
void solve(int i, int p, vector<int>& res) {
  for (auto t : conns[i]) {
    if (t != p && siz[t] % 2 == 0) solve(t, i, res);
  }
  res.push_back(i);
  for (auto t : conns[i]) {
    if (t != p && siz[t] % 2 == 1) solve(t, i, res);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p;
    --p;
    if (p != -1) {
      conns[i].push_back(p);
      conns[p].push_back(i);
    }
  }
  if (n % 2 == 0) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    vector<int> res;
    dfs(0, 0);
    solve(0, 0, res);
    for (auto i : res) cout << i + 1 << '\n';
  }
}