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;
const int N = 2e5 + 5;
int n, root;
bool visit[N];
int deg[N];
int h[N];
vector<int> G[N];
vector<int> res;
void dfs(int u, int p) {
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    if (v == p) continue;
    h[v] = h[u] + 1;
    dfs(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  if (n % 2 == 0) return cout << "NO" << endl, 0;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p;
    if (p == 0)
      root = i;
    else {
      G[p].push_back(i);
      G[i].push_back(p);
      ++deg[i];
      ++deg[p];
    }
  }
  dfs(root, root);
  priority_queue<pair<int, int> > pq;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] % 2 == 0) pq.push(make_pair(h[i], i));
  }
  while (!pq.empty()) {
    pair<int, int> u = pq.top();
    pq.pop();
    if (visit[u.second] || deg[u.second] & 1) continue;
    deg[u.second] = 0;
    visit[u.second] = true;
    res.push_back(u.second);
    for (int i = 0; i < G[u.second].size(); ++i) {
      int v = G[u.second][i];
      if (deg[v] == 0) continue;
      --deg[v];
      if (deg[v] % 2 == 0) pq.push(make_pair(h[v], v));
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; ++i) cout << res[i] << endl;
}
