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>
const long long INF = 1000000000;
const long long N = 300005;
const long long MOD = 1000000007;
const long long MAXN = 100005;
const long long rootval = 319;
using namespace std;
vector<long long> adj[N];
vector<long long> v;
long long dfs(long long u, long long par) {
  long long curr = 1;
  for (auto v : adj[u]) {
    if (v == par) {
      continue;
    }
    curr = curr + dfs(v, u);
  }
  if (curr % 2 == 0) {
    cout << u + 1 << endl;
  } else {
    v.push_back(u + 1);
  }
  return curr;
}
int main() {
  long long n;
  cin >> n;
  long long p[n + 1];
  for (long long i = 0; i < n; i++) {
    cin >> p[i];
    if (p[i] == 0) {
      continue;
    }
    adj[p[i] - 1].push_back(i);
    adj[i].push_back(p[i] - 1);
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    dfs(0, -1);
    if (!v.empty()) {
      reverse(v.begin(), v.end());
      for (long long i = 0; i < v.size(); i++) {
        cout << v[i] << endl;
      }
    }
  }
}
