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;
const int MAXN = 200100;
vector<int> ed[MAXN];
int sz[MAXN];
void dfs1(int cur, int par = -1) {
  vector<int> &v = ed[cur];
  sz[cur] = 1;
  for (int i = 0; i < v.size(); ++i) {
    int nxt = v[i];
    if (nxt == par) continue;
    dfs1(nxt, cur);
    sz[cur] += sz[nxt];
  }
}
vector<int> ans;
void dfs2(int cur, int par = -1) {
  vector<int> &v = ed[cur];
  for (int i = 0; i < v.size(); ++i) {
    int nxt = v[i];
    if (nxt == par) continue;
    if (sz[nxt] % 2 == 0) {
      dfs2(nxt, cur);
    }
  }
  ans.push_back(cur);
  for (int i = 0; i < v.size(); ++i) {
    int nxt = v[i];
    if (nxt == par) continue;
    if (sz[nxt] % 2 == 1) {
      dfs2(nxt, cur);
    }
  }
}
int n;
int main() {
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO"
         << "\n";
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p;
    if (p) {
      ed[i].push_back(p);
      ed[p].push_back(i);
    }
  }
  dfs1(1);
  dfs2(1);
  cout << "YES\n";
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << "\n";
  }
}
