Problem: 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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
int n, root;
int sz[N + 5];
vector<int> v[N + 5];
void dfs1(int nod) {
  int ans = 0;
  for (auto nou : v[nod]) {
    dfs1(nou);
    ans += sz[nou];
  }
  sz[nod] = ans + 1;
}
void dfs(int nod) {
  for (auto nou : v[nod])
    if (sz[nou] % 2 == 0) dfs(nou);
  cout << nod << "\n";
  for (auto nou : v[nod])
    if (sz[nou] % 2 == 1) dfs(nou);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int nod = 1; nod <= n; nod++) {
    int t;
    cin >> t;
    if (t == 0)
      root = nod;
    else
      v[t].push_back(nod);
  }
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  dfs1(root);
  dfs(root);
  return 0;
}