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;
const int N = 2e5 + 7;
vector<int> adj[N];
int ans[N], a[N], idx[N];
int dfs(int u, int p) {
  int cnt = 0;
  for (int v : adj[u])
    if (v != p) {
      cnt += dfs(v, u);
    }
  return ans[u] = (int(adj[u].size()) - cnt) % 2 == 0;
}
void DFS(int u, int p) {
  if (ans[u])
    a[u] = a[p] - 1;
  else
    a[u] = a[p] + 1;
  for (int v : adj[u])
    if (v != p) {
      DFS(v, u);
    }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int j;
    cin >> j;
    if (j) {
      adj[i].push_back(j);
      adj[j].push_back(i);
    }
  }
  int root = -1;
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() != 1) root = i;
  }
  if (root == -1) {
    cout << "NO" << endl;
    return 0;
  }
  if (dfs(root, 0)) {
    cout << "YES" << endl;
    DFS(root, 0);
    vector<pair<int, int> > xx;
    for (int i = 1; i <= n; i++) xx.push_back({a[i], i});
    sort(xx.begin(), xx.end());
    for (int i = 1; i <= n; i++) cout << xx[i - 1].second << endl;
  } else
    cout << "NO" << endl;
}