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 = 2 * 100000 + 100;
vector<int> e[N];
bool mark[N], del[N];
int n, cnt, par[N];
void destruct(int v) {
  del[v] = true;
  cout << v + 1 << endl;
  for (int i = 0; i < e[v].size(); i++) {
    int u = e[v][i];
    if (u != par[v] && !del[u]) destruct(u);
  }
}
int dfs(int v) {
  mark[v] = true;
  int s = 0, d = 0;
  for (int i = 0; i < e[v].size(); i++) {
    int u = e[v][i];
    if (!mark[u]) {
      par[u] = v;
      int x = dfs(u);
      s += x;
    }
  }
  if (s % 2) destruct(v);
  return s + 1;
}
int main() {
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    int v;
    cin >> v;
    if (v == 0) continue;
    v--;
    e[v].push_back(i);
    e[i].push_back(v);
  }
  dfs(0);
  destruct(0);
}
