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 = 3e5 + 5;
int n, x, a[N];
vector<int> g[N], q1, q2;
void dfs(int v, int pr = 0) {
  for (int to : g[v]) {
    if (to != pr) {
      dfs(to, v);
    }
  }
  if (a[v] % 2 == 0) {
    q1.push_back(v);
    a[pr]--;
  } else {
    q2.push_back(v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x) {
      a[x]++;
      a[i]++;
      g[i].push_back(x);
      g[x].push_back(i);
    }
  }
  dfs(1);
  reverse(q2.begin(), q2.end());
  for (auto it : q2) {
    q1.push_back(it);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = g[i].size();
  }
  for (auto it : q1) {
    if (a[it] & 1) {
      return cout << "NO", 0;
    }
    for (auto to : g[it]) {
      a[to]--;
    }
  }
  cout << "YES\n";
  for (auto it : q1) {
    cout << it << "\n";
  }
}
