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 deg[200040], erased[200500], depth[200040];
void dfs(int x, int d, vector<vector<int>>& adj, int pa) {
  for (auto u : adj[x])
    if (u != pa) dfs(u, d + 1, adj, x);
  depth[x] = d;
}
int main() {
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  int root;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p == 0) {
      root = i;
      continue;
    }
    adj[p].push_back(i);
    adj[i].push_back(p);
    deg[i]++;
    deg[p]++;
  }
  dfs(root, 0, adj, 0);
  set<pair<int, int>> S;
  for (int i = 1; i <= n; i++) {
    if (deg[i] % 2 == 0) S.insert({-depth[i], i});
  }
  vector<int> res;
  while (!S.empty()) {
    int x = (*S.begin()).second;
    S.erase({-depth[x], x});
    erased[x] = 1;
    for (auto u : adj[x]) {
      if (erased[u]) continue;
      deg[u]--;
      if (deg[u] % 2 == 0) {
        S.insert({-depth[u], u});
      } else
        S.erase({-depth[u], u});
    }
    res.push_back(x);
  }
  if (res.size() != n) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (auto u : res) {
    cout << u << endl;
  }
  return 0;
}