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;
int n, p[200003], edge[200003], cnt[200003], level[200003];
vector<int> ve, g[200003];
bool sudah[200003], ya[200003], yy[200003];
void dfs(int now) {
  ya[now] = true;
  ve.push_back(now);
  for (int i : g[now]) {
    if (ya[i]) continue;
    dfs(i);
  }
}
void aa(int now) {
  yy[now] = true;
  for (int i : g[now]) {
    if (yy[i]) continue;
    level[i] = level[now] + 1;
    aa(i);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    g[p[i]].push_back(i);
    if (p[i] == 0) continue;
    edge[i]++;
    edge[p[i]]++;
    cnt[p[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == 0) aa(i);
  }
  priority_queue<pair<int, int>> q;
  for (int i = 1; i <= n; i++) {
    if (!cnt[i]) {
      q.push({level[i], i});
      sudah[i] = true;
    }
  }
  while (!q.empty()) {
    auto now = q.top();
    q.pop();
    if (edge[now.second] % 2 == 0) {
      if (!ya[now.second]) {
        dfs(now.second);
        edge[p[now.second]]--;
      }
    }
    if (!sudah[p[now.second]] && p[now.second] != 0) {
      q.push({level[p[now.second]], p[now.second]});
      sudah[p[now.second]] = true;
    }
  }
  if (ve.size() != n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i : ve) cout << i << endl;
}