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;
std::vector<long long int> aarr;
bool visited[200005];
std::vector<long long int> v[200005];
long long int root;
void dfs2(long long int s) {
  if (visited[s]) return;
  visited[s] = true;
  aarr.push_back(s);
  for (long long int i = 0; i < v[s].size(); ++i) {
    dfs2(v[s][i]);
  }
}
void dfs(long long int s) {
  long long int deg = 0;
  if (s != root) deg++;
  for (long long int i = 0; i < v[s].size(); ++i) {
    dfs(v[s][i]);
    if (!visited[v[s][i]]) deg++;
  }
  if (deg % 2 == 0) dfs2(s);
}
int main() {
  long long int n, a, b, c, t;
  cin >> n;
  long long int p[n + 1], cnt = 0;
  memset(visited, 0, sizeof(visited));
  for (long long int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] != 0) {
      cnt++;
      v[p[i]].push_back(i);
    } else {
      root = i;
    }
  }
  if (cnt % 2 != 0) {
    cout << "NO\n";
    return 0;
  }
  dfs(root);
  cout << "YES\n";
  for (long long int i = 0; i < aarr.size(); i++) {
    cout << aarr[i] << endl;
  }
}
