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;
vector<int> v[200010], g;
int vis[200010];
int root, sum;
void Check(int r) {
  if (vis[r]) return;
  vis[r] = 1;
  g.push_back(r);
  int sz = v[r].size();
  for (int i = 0; i < sz; i++) Check(v[r][i]);
}
void Dfs(int r) {
  int sz = v[r].size();
  int deg = 0;
  if (r != root) deg++;
  for (int i = 0; i < sz; i++) {
    Dfs(v[r][i]);
    if (!vis[v[r][i]]) deg++;
  }
  if (!(deg & 1)) Check(r);
}
int main() {
  int n, b;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> b;
    if (b == 0)
      root = i;
    else
      v[b].push_back(i);
  }
  Dfs(root);
  int sz = g.size();
  if (sz == n) {
    cout << "YES" << endl;
    for (int i = 0; i < sz; i++) cout << g[i] << endl;
  } else
    cout << "NO" << endl;
  return 0;
}