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.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e5 + 9, MOD = 1e9 + 7;
int n, dp[maxN];
vector<int> adj[maxN];
vector<int> ans;
void dfs(int nd, int an) {
  dp[nd] = 1;
  for (int sn : adj[nd]) {
    if (sn == an) continue;
    dfs(sn, nd);
    dp[nd] ^= dp[sn];
  }
}
void cons(int nd, int an) {
  for (int sn : adj[nd]) {
    if (sn != an && dp[sn] == 0) {
      cons(sn, nd);
    }
  }
  ans.push_back(nd + 1);
  for (int sn : adj[nd]) {
    if (sn != an && dp[sn] == 1) {
      cons(sn, nd);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0, inp; i < n; i++) {
    cin >> inp;
    inp--;
    if (inp != -1) {
      adj[i].push_back(inp);
      adj[inp].push_back(i);
    }
  }
  dfs(0, -1);
  if (dp[0] == 0) {
    cout << "NO\n";
    return 0;
  }
  dp[0] = 0;
  cons(0, -1);
  cout << "YES\n";
  for (int el : ans) cout << el << '\n';
}