Problem: 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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> mg[200005];
int otac[200005], marka[200005], st[200005];
void brisanje(int cvor) {
  int i, x;
  marka[cvor] = 3;
  cout << cvor << endl;
  for (i = 0; i < mg[cvor].size(); i++) {
    x = mg[cvor][i];
    if (marka[x] == 1) st[x]--;
    if (marka[x] == 1 && x != otac[cvor]) brisanje(x);
  }
}
void dfs(int cvor) {
  int i, x;
  marka[cvor] = 1;
  for (i = 0; i < mg[cvor].size(); i++) {
    x = mg[cvor][i];
    if (marka[x] == 0) {
      otac[x] = cvor;
      dfs(x);
    }
  }
  if (st[cvor] % 2 == 0) brisanje(cvor);
}
int main() {
  int n, i, j, x;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    if (x) {
      mg[i].push_back(x);
      mg[x].push_back(i);
      st[i]++;
      st[x]++;
    }
  }
  if (n % 2 == 0)
    cout << "NO";
  else {
    cout << "YES" << endl;
    dfs(1);
  }
  return 0;
}