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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int INF = 2147483647;
const int mod = 1000000007;
int f[200011];
int a[200011];
vector<int> v[200011];
vector<int> ans;
void dfs2(int node, int fa) {
  a[node] = 0;
  ans.push_back(node);
  for (int i = 0; i < (v[node].size()); i++) {
    if (v[node][i] == fa) continue;
    if (a[v[node][i]] == 1) dfs2(v[node][i], node);
  }
}
void dfs(int node, int fa) {
  if (v[node].size() == 1 && node != 1) {
    a[node] = 1;
    return;
  }
  int sum = 0;
  for (int i = 0; i < (v[node].size()); i++) {
    if (v[node][i] == fa) continue;
    dfs(v[node][i], node);
    sum += a[v[node][i]];
  }
  if (node != 1) sum++;
  if (sum % 2 == 1)
    a[node] = 1;
  else
    dfs2(node, fa);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= (n); i++) {
    cin >> f[i];
    if (!f[i]) continue;
    v[i].push_back(f[i]);
    v[f[i]].push_back(i);
  }
  dfs(1, -1);
  if (a[1] == 0) {
    cout << "YES" << '\n';
    for (int i = 0; i < (ans.size()); i++) cout << ans[i] << '\n';
  } else
    cout << "NO" << '\n';
  return 0;
}
