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 P = 1e9 + 9;
const int MAXN = 200200;
vector<int> g[MAXN];
int del[MAXN];
int deg[MAXN];
void dfs1(int v, int p = -1) {
  del[v] = 1;
  cout << v + 1 << endl;
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    if (!del[u]) {
      dfs1(u, v);
    }
  }
}
void dfs(int v, int p = -1) {
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    dfs(u, v);
  }
  int deg = (int)(g[v]).size();
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    if (del[u]) {
      --deg;
    }
  }
  if (deg % 2 == 0) {
    dfs1(v, p);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    --p[i];
    if (p[i] != -1) {
      g[p[i]].push_back(i);
      g[i].push_back(p[i]);
    }
  }
  if (n == 1) {
    cout << "YES" << endl;
    cout << 1 << endl;
    return 0;
  }
  int e = 0;
  for (int i = 0; i < n; ++i) {
    if ((int)(g[i]).size() % 2 == 0) {
      ++e;
    }
  }
  if (e % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  dfs(0);
  return 0;
}
