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.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int p[200002];
vector<int> ke[200002];
bool mark[200002];
bool dfs(int u, int p) {
  int cnt = 0;
  for (int v : ke[u]) {
    if (v == p) continue;
    if (dfs(v, u)) cnt++;
  }
  if (cnt % 2 == 0) {
    return mark[u] = true;
  }
  return mark[u] = false;
}
void dfs2(int u, int p) {
  for (int v : ke[u]) {
    if (v == p) continue;
    if (!mark[v]) {
      dfs2(v, u);
    }
  }
  cout << u << "\n";
  for (int v : ke[u]) {
    if (v == p) continue;
    if (mark[v]) {
      dfs2(v, u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i]) {
      ke[i].push_back(p[i]);
      ke[p[i]].push_back(i);
    }
  }
  if (dfs(1, 1)) {
    cout << "YES\n";
    dfs2(1, 1);
  } else {
    cout << "NO";
  }
}