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 MOD = 1e9 + 7;
int n;
vector<vector<int>> ed;
vector<int> subt, answer;
void init() {
  ed = vector<vector<int>>(n + 1);
  subt = vector<int>(n + 1);
  answer = vector<int>(n + 1);
}
int dfs(int fa, int u) {
  int val = 1;
  for (auto i : ed[u]) {
    if (i != fa) {
      val = val + dfs(u, i);
    }
  }
  subt[u] = val;
  return val;
}
void solve(int fa, int u) {
  for (auto i : ed[u]) {
    if (i != fa && subt[i] % 2 == 0) solve(u, i);
  }
  cout << u << "\n";
  for (auto i : ed[u]) {
    if (i != fa && subt[i] % 2) solve(u, i);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  init();
  vector<int> inp(n);
  for (int i = 0; i < n; i++) {
    cin >> inp[i];
  }
  for (int i = 0; i < n; i++) {
    if (inp[i]) {
      int u = i + 1, v = inp[i];
      ed[u].push_back(v);
      ed[v].push_back(u);
    }
  }
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  dfs(0, 1);
  cout << "YES\n";
  solve(0, 1);
  return 0;
}
