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 d[1000006], n, root;
vector<int> e[1000006];
void dfs1(int u) {
  d[u] = 1;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    dfs1(v);
    d[u] += d[v];
  }
  return;
}
void dfs2(int u) {
  for (int i = 0; i < e[u].size(); i++)
    if (d[e[u][i]] % 2 == 0) dfs2(e[u][i]);
  cout << u << endl;
  for (int i = 0; i < e[u].size(); i++)
    if (d[e[u][i]] % 2) dfs2(e[u][i]);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  if (n % 2 == 0) return cout << "NO", 0;
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x == 0)
      root = i;
    else
      e[x].push_back(i);
  }
  dfs1(root);
  dfs2(root);
  return 0;
}