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;
int n;
int sz[maxN];
vector<int> g[maxN], ans;
void ReadInput() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p != 0) {
      g[i].push_back(p);
      g[p].push_back(i);
    }
  }
}
int dfs1(int u, int p) {
  sz[u] = 1;
  for (int v : g[u])
    if (v != p) sz[u] += dfs1(v, u);
  return sz[u];
}
void dfs2(int u, int p) {
  for (int v : g[u])
    if (v != p && sz[v] % 2 == 0) dfs2(v, u);
  ans.push_back(u);
  for (int v : g[u])
    if (v != p && sz[v] % 2 != 0) dfs2(v, u);
}
void Solve() {
  if (n % 2 == 0) {
    cout << "NO";
    return;
  }
  dfs1(1, 0);
  dfs2(1, 0);
  cout << "YES\n";
  for (int it : ans) cout << it << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ReadInput();
  Solve();
}
