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;
vector<int> D[200000], T;
int S[200000];
void dfsS(int d, int p) {
  S[d] = 1;
  for (int i = 0; i < D[d].size(); i++) {
    if (D[d][i] != p) {
      dfsS(D[d][i], d);
      S[d] += S[D[d][i]];
    }
  }
  return;
}
void dfs(int d, int p) {
  int A;
  for (int i = 0; i < D[d].size(); i++) {
    A = D[d][i];
    if (A != p && S[A] % 2 == 0) {
      dfs(A, d);
    }
  }
  cout << d + 1 << "\n";
  for (int i = 0; i < D[d].size(); i++) {
    A = D[d][i];
    if (A != p && S[A] % 2 == 1) {
      dfs(A, d);
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  int n, a;
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES"
       << "\n";
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a != 0) {
      a--;
      D[a].push_back(i);
      D[i].push_back(a);
    }
    S[i] = 0;
  }
  T.push_back(0);
  dfsS(0, 0);
  dfs(0, -1);
  return 0;
}
