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, p[200001], count_child[200001];
vector<int> adj[200001], comb;
bool visited[200001], del[200001], visited2[200001];
void setup() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int temp;
    cin >> temp;
    if (temp) {
      adj[temp].push_back(i);
      adj[i].push_back(temp);
    }
  }
}
void dfs2(int t) {
  visited2[t] = true;
  if (!del[t]) comb.push_back(t);
  for (int i : adj[t]) {
    if (!visited2[i] && i != p[t]) {
      dfs2(i);
    }
  }
}
void dfs(int t) {
  visited[t] = true;
  for (int i : adj[t]) {
    if (!visited[i]) {
      p[i] = t;
      dfs(i);
      if (!del[i]) count_child[t]++;
    }
  }
  if (count_child[t] & 1 || (t == 1 && count_child[t] % 2 == 0)) {
    del[t] = true;
    comb.push_back(t);
    dfs2(t);
  }
}
void process() {
  dfs(1);
  if (count_child[1] & 1) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (int i : comb) cout << i << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  setup();
  process();
  return 0;
}