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;
void JIZZ(string output = "") {
  cout << output;
  exit(0);
}
const long double PI = 3.14159265358979323846264338327950288;
const long double eps = 1e-13;
const long long mod = 1e9 + 7;
vector<int> G[200005];
int p[200005], dep[200005], deg[200005];
bitset<200005> inq, del;
vector<int> ans;
void go(int now, int pa) {
  if (del[now]) return;
  ans.push_back(now);
  del[now] = 1;
  for (int i : G[now]) {
    if (i != pa) go(i, now);
  }
}
void dfs(int now, int pa) {
  ;
  ;
  for (int i : G[now]) {
    if (i == pa) continue;
    dfs(i, now);
  }
  if (deg[now] % 2 == 0) {
    go(now, pa);
    --deg[pa];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  if (n == 1) exit((cout << "YES\n1\n", 0));
  if (n == 2) exit((cout << "NO" << endl, 0));
  int root = 0;
  for (int i = 1; i <= n; ++i) {
    int pa;
    cin >> pa;
    if (pa)
      G[pa].push_back(i), G[i].push_back(pa);
    else
      root = i;
  }
  for (int i = 1; i <= n; ++i) deg[i] = G[i].size();
  dfs(root, root);
  ;
  ;
  if (ans.size() != n) exit((cout << "NO" << endl, 0));
  cout << "YES" << endl;
  for (int i : ans) cout << i << '\n';
}