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> Tree[200005], ans;
stack<int> S;
int root, a, x, odw[200005];
bool DFS(int n) {
  odw[n] = true;
  int cnt = 1;
  if (n == root) cnt--;
  for (int i = 0; i < Tree[n].size(); i++)
    if (!odw[Tree[n][i]]) cnt += (int)DFS(Tree[n][i]);
  if (cnt % 2)
    S.push(n);
  else {
    ans.push_back(n);
    return false;
  }
  return true;
}
int main() {
  cin >> a;
  for (int i = 1; i <= a; i++) {
    cin >> x;
    if (!x) {
      root = i;
      continue;
    }
    Tree[i].push_back(x);
    Tree[x].push_back(i);
  }
  if (DFS(root)) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << "\n";
  while (!S.empty()) {
    cout << S.top() << "\n";
    S.pop();
  }
  return 0;
}
