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;
int p[200002], d[200002];
vector<int> ch[200002];
vector<int> ans, buf;
void bfs(int pos) {
  for (int c : ch[pos]) bfs(c);
  if (d[pos] & 1) {
    buf.push_back(pos);
  } else {
    ans.push_back(pos);
    d[p[pos]]--;
  }
}
int main() {
  int n;
  cin >> n;
  int root = 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] == 0) {
      root = i;
    } else {
      ch[p[i]].push_back(i);
    }
    d[i]++;
    d[p[i]]++;
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  bfs(root);
  for (int a : ans) cout << a << endl;
  reverse(buf.begin(), buf.end());
  for (int b : buf) cout << b << endl;
  return 0;
}