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 Nmax = 2e5 + 2, inf = 1e9 + 7, base = 1e9;
vector<int> g[Nmax];
int cnt[Nmax];
bool killed[Nmax];
void dfs(int v, int p = -1, bool go = true) {
  if (go)
    for (int w : g[v]) {
      if (w == p) continue;
      if (killed[w]) continue;
      dfs(w, v);
    }
  if (!(cnt[v] & 1)) {
    killed[v] = true;
    cout << v + 1 << '\n';
    for (int w : g[v]) {
      cnt[w]--;
      if (w != p && !killed[w]) dfs(w, v, false);
    }
  }
}
int main() {
  int n, hV;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> hV;
    if (hV != 0) {
      g[i].push_back(--hV);
      g[hV].push_back(i);
    }
  }
  for (int i = 0; i < n; ++i) {
    cnt[i] = g[i].size();
  }
  if (n & 1) {
    cout << "YES\n";
    dfs(0);
    for (int i = 0; i < n; ++i) {
      if (!killed[i]) cout << i + 1 << '\n';
    }
  } else
    cout << "NO\n";
  return 0;
}
