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