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 n;
vector<vector<int> > v;
int root;
vector<int> deleted;
vector<int> taken;
void dd(int s) {
  if (deleted[s]) return;
  deleted[s] = 1;
  taken.push_back(s);
  for (int i = 0; i < v[s].size(); i++) dd(v[s][i]);
}
bool dfs(int x) {
  int edges = 0;
  if (x != root) edges++;
  for (int i = 0; i < v[x].size(); i++) {
    if (!dfs(v[x][i]) && !deleted[v[x][i]]) edges++;
  }
  if (edges % 2 == 0) {
    dd(x);
    return 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  v.resize(n + 1);
  deleted.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x != 0) {
      v[x].push_back(i);
    } else {
      root = i;
    }
  }
  if (dfs(root) == 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 0; i < taken.size(); i++) {
    cout << taken[i] << endl;
  }
  return 0;
}