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 = 234567;
int n;
vector<int> e[N];
int f[N];
void dfs(int u, int p) {
  int go_before_parent = 0;
  int go_after_parent = 0;
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == p) {
      continue;
    }
    dfs(v, u);
    if (f[v] == 0) {
      ++go_before_parent;
    } else {
      ++go_after_parent;
    }
  }
  int son_degree = go_after_parent;
  if (son_degree & 1) {
    f[u] = 0;
  } else {
    f[u] = 1;
  }
}
void print_solution(int u, int p) {
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == p) {
      continue;
    }
    if (f[v] == 0) {
      print_solution(v, u);
    }
  }
  cout << u << endl;
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == p) {
      continue;
    }
    if (f[v] == 1) {
      print_solution(v, u);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1, j; i <= n; ++i) {
    cin >> j;
    if (j != 0) {
      e[i].push_back(j);
      e[j].push_back(i);
    }
  }
  dfs(1, 0);
  if (f[1] == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  print_solution(1, 0);
  return 0;
}
