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 s[200010];
int p[200010];
bool odw[200010];
vector<int> g[200010];
queue<int> kolej;
int n, w, k;
void dfs(int v, int pop = 0) {
  for (unsigned int i = 0; i < g[v].size(); i++)
    if (g[v][i] != pop) dfs(g[v][i], v);
}
void dfs3(int v) {
  kolej.push(v);
  odw[v] = 1;
  for (unsigned int i = 0; i < g[v].size(); i++) {
    w = g[v][i];
    s[w]--;
    if (w == p[v]) continue;
    if (!odw[w]) dfs3(w);
  }
}
void dfs2(int v) {
  for (unsigned int i = 0; i < g[v].size(); i++)
    if (g[v][i] != p[v]) dfs2(g[v][i]);
  if (!(s[v] & 1)) dfs3(v);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] > 0) {
      g[i].push_back(p[i]);
      g[p[i]].push_back(i);
    } else
      k = i;
  }
  for (int i = 1; i <= n; i++) s[i] = g[i].size();
  dfs(k);
  dfs2(k);
  if (kolej.size() != n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  while (!kolej.empty()) {
    cout << kolej.front() << "\n";
    kolej.pop();
  }
  return 0;
}
