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> v[200005], sol;
queue<int> q;
int deg[200005], t[200005];
bool viz[200005];
void dfs(int i);
void afiseaza(int i);
int main() {
  int n, i, a, wait, root;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    cin >> a;
    t[i] = a;
    if (a) {
      v[i].push_back(a);
      v[a].push_back(i);
      ++deg[a];
      ++deg[i];
    } else {
      root = i;
    }
  }
  dfs(root);
  if (sol.size() < n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (auto it = sol.begin(); it != sol.end(); ++it) {
    cout << *it << '\n';
  }
}
void dfs(int i) {
  for (auto it = v[i].begin(); it != v[i].end(); ++it) {
    if (*it != t[i]) {
      dfs(*it);
    }
  }
  if (deg[i] % 2 == 0) {
    afiseaza(i);
    --deg[t[i]];
  }
}
void afiseaza(int i) {
  viz[i] = 1;
  sol.push_back(i);
  for (auto it = v[i].begin(); it != v[i].end(); ++it) {
    if (*it != t[i] && !viz[*it]) {
      afiseaza(*it);
    }
  }
}