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;
vector<int> v[200005];
vector<int> nn;
int s[200005], t = -1;
void dfss(int x) {
  if (s[x]) return;
  s[x] = 1;
  nn.push_back(x);
  for (int i = 0; i < v[x].size(); i++) dfss(v[x][i]);
  return;
}
void dfs(int x) {
  int ss = 0;
  if (x != t) ss++;
  for (int i = 0; i < v[x].size(); i++) {
    dfs(v[x][i]);
    if (s[v[x][i]] == 0) ss++;
  }
  if (ss % 2 == 0) {
    dfss(x);
  }
  return;
}
int main() {
  int n;
  cin >> n;
  int j;
  for (int i = 1; i <= n; i++) {
    cin >> j;
    if (j == 0) {
      t = i;
      continue;
    }
    v[j].push_back(i);
  }
  dfs(t);
  if (nn.size() != n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (vector<int>::iterator i = nn.begin(); i != nn.end(); i++) {
    cout << *i << endl;
  }
}
