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;
int vis[500045];
std::vector<int> a[500045];
std::vector<int> order;
void dfs1(int pos, int par) {
  if (vis[pos]) return;
  vis[pos] = 1;
  order.push_back(pos);
  for (int i = 0; i < a[pos].size(); i++) {
    if (a[pos][i] != par) {
      dfs1(a[pos][i], pos);
    }
  }
}
int dfs(int pos, int par) {
  int deg = 1;
  if (par == 0) {
    deg = 0;
  }
  for (int i = 0; i < a[pos].size(); i++) {
    if (a[pos][i] == par) continue;
    deg += dfs(a[pos][i], pos);
  }
  if (deg % 2 == 0) {
    a[pos].push_back(par);
    dfs1(pos, par);
    return 0;
  } else {
    return 1;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int p[n];
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] == 0) continue;
    a[i].push_back(p[i]);
    a[p[i]].push_back(i);
  }
  dfs(1, 0);
  if (order.size() != n) {
    cout << "NO\n";
    return 0;
  } else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      cout << order[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}
