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;
int const N = 2e5 + 10;
int n, s, subtree[N], deg[N], par[N];
vector<int> adj[N];
bool mark[N];
void read_input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a) {
      a--;
      adj[i].push_back(a);
      adj[a].push_back(i);
      deg[a]++, deg[i]++;
    }
  }
}
void dstr(int v) {
  if (!mark[v]) {
    cout << v + 1 << '\n';
    mark[v] = true;
  }
  for (int q : adj[v])
    if (!mark[q] && q != par[v]) dstr(q);
}
void dfs(int s) {
  subtree[s] = 1;
  for (int u : adj[s])
    if (u != par[s]) {
      par[u] = s;
      dfs(u);
      subtree[s] += subtree[u];
    }
  if (!(subtree[s] % 2)) dstr(s);
}
void solve() {
  if (n % 2 == 0) {
    cout << "NO";
    exit(0);
  }
  for (int i = 0; i < n; i++)
    if (deg[i] == 1) s = i;
  cout << "YES\n";
  if (n == 1)
    cout << 1;
  else {
    par[s] = -1;
    dfs(s);
    cout << s + 1;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input();
  solve();
}