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;
vector<int> adj[200001];
pair<int, int> below[200001];
int even[200001];
int dfs(int src) {
  if ((int)adj[src].size() == 0) return even[src] = 1;
  int ok = 0, n = 0;
  for (int j = 0; j < (int)adj[src].size(); j++) {
    int cur = adj[src][j];
    if (dfs(cur))
      n++;
    else
      ok++;
  }
  below[src] = make_pair(ok, n);
  if ((n + 1) % 2 == 0)
    return even[src] = 0;
  else
    return even[src] = 1;
}
void dfs_trace(int src, vector<int>& ans) {
  if ((int)adj[src].size() == 0) {
    ans.push_back(src);
    return;
  }
  for (int j = 0; j < (int)adj[src].size(); j++) {
    int cur = adj[src][j];
    if (even[cur] == 0) dfs_trace(cur, ans);
  }
  ans.push_back(src);
  for (int j = 0; j < (int)adj[src].size(); j++) {
    int cur = adj[src][j];
    if (even[cur] == 1) dfs_trace(cur, ans);
  }
}
bool solve(vector<int>& ans, int root) {
  int x = dfs(root);
  if (x == 0) return false;
  dfs_trace(root, ans);
  return true;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, root;
  cin >> n;
  for (int j = 1; j <= n; j++) {
    int par;
    cin >> par;
    if (par == 0) root = j;
    adj[par].push_back(j);
  }
  if (n == 1) {
    cout << "YES\n";
    cout << 1 << endl;
    return 0;
  }
  vector<int> ans;
  if (solve(ans, root)) {
    cout << "YES\n";
    for (int j = 0; j < (int)ans.size(); j++) cout << ans[j] << endl;
  } else
    cout << "NO\n";
  return 0;
}