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;
using GRAPH = std::vector<std::vector<int>>;
GRAPH G;
vector<int> P;
vector<int> D;
vector<int> V;
void dfs(int u) {
  for (auto v : G[u]) {
    dfs(v);
  }
  V.push_back(u);
}
int main() {
  int n{};
  cin >> n;
  G.resize(n + 1);
  P.resize(n + 1, 0);
  D.resize(n + 1, 0);
  int root{};
  for (int i = 1; i <= n; ++i) {
    cin >> P[i];
    if (P[i] == 0) {
      root = i;
    } else {
      ++D[i];
    }
    ++D[P[i]];
    G[P[i]].push_back(i);
  }
  dfs(root);
  vector<int> ans;
  vector<int> e;
  for (auto v : V) {
    if (D[v] % 2 == 1) {
      if (v == root) {
        cout << "NO\n";
        return 0;
      }
      e.push_back(v);
    } else {
      ans.push_back(v);
      --D[P[v]];
    }
  }
  cout << "YES" << endl;
  for (auto u : ans) {
    cout << u << endl;
  }
  for (int i = e.size() - 1; i >= 0; --i) {
    cout << e[i] << endl;
  }
}