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.

<image>
#include <bits/stdc++.h>
using namespace std;
int sz[200010];
int vis[200010];
int p[200010];
vector<int> son[200010];
void del(int x) {
  vis[x] = 1;
  cout << x << endl;
  for (int i = 0; i < son[x].size(); i++)
    if (!vis[son[x][i]]) del(son[x][i]);
}
void dfs(int x) {
  sz[x] = 1;
  for (int i = 0; i < son[x].size(); i++)
    dfs(son[x][i]), sz[x] += sz[son[x][i]];
  if (!(sz[x] & 1)) del(x);
}
int main() {
  int n;
  int root;
  cin >> n;
  if (n & 1)
    cout << "YES" << endl;
  else
    return cout << "NO" << endl, 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i])
      son[p[i]].push_back(i);
    else
      root = i;
  }
  dfs(root);
  del(root);
  return 0;
}
