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 MAXN = 2e5 + 5;
const int INF = 1e9 + 5;
int child[MAXN];
bitset<MAXN> done;
vector<int> rta;
vector<int> G[MAXN];
int dfs1(int st) {
  done[st] = true;
  int sum = 0;
  for (auto &i : G[st])
    if (!done[i]) sum += dfs1(i);
  return child[st] = sum + 1;
}
void dfs2(int st) {
  done[st] = true;
  for (auto &i : G[st])
    if (!done[i] && !(child[i] & 1)) dfs2(i);
  rta.push_back(st);
  for (auto &i : G[st])
    if (!done[i] && (child[i] & 1)) dfs2(i);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < int(n); i++) {
    int x;
    cin >> x;
    x--;
    if (x >= 0) G[i].push_back(x), G[x].push_back(i);
  }
  if (!(n & 1)) return cout << "NO", 0;
  dfs1(0);
  done.reset();
  dfs2(0);
  cout << "YES\n";
  for (auto &i : rta) cout << i + 1 << '\n';
  return 0;
}
