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<long long> adj[200005];
bool vis[200005] = {false};
long long degree[200005] = {0};
vector<long long> ans;
long long dfs(long long s) {
  vis[s] = true;
  long long i, ans = 0;
  for (i = 0; i < adj[s].size(); i++) {
    if (!vis[adj[s][i]]) ans += dfs(adj[s][i]);
  }
  degree[s] = ans + 1;
  return degree[s];
}
void dfs1(long long s) {
  vis[s] = true;
  long long i;
  for (i = 0; i < adj[s].size(); i++) {
    if (!vis[adj[s][i]] && degree[adj[s][i]] % 2 == 0) dfs1(adj[s][i]);
  }
  ans.push_back(s);
  for (i = 0; i < adj[s].size(); i++) {
    if (!vis[adj[s][i]] && degree[adj[s][i]] % 2 == 1) dfs1(adj[s][i]);
  }
}
int main() {
  long long n, i, j, temp;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> temp;
    if (temp == 0) continue;
    adj[i].push_back(temp);
    adj[temp].push_back(i);
  }
  if (n % 2 == 0) {
    cout << "NO"
         << "\n";
    return 0;
  }
  dfs(1);
  for (i = 1; i <= n; i++) vis[i] = false;
  dfs1(1);
  cout << "YES"
       << "\n";
  for (i = 0; i < ans.size(); i++) cout << ans[i] << "\n";
}