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 + 99;
int n, x, rt, mark[N], deg[N], dist[N], par[N];
set<pair<int, int> > s;
vector<int> ans, g[N];
void update(int x) {
  deg[x]--;
  if (deg[x] % 2)
    s.erase(make_pair(dist[x], x));
  else
    s.insert(make_pair(dist[x], x));
}
void dfs(int x, int dis) {
  dist[x] = dis;
  deg[x] = (g[x].size() + (x != rt));
  if (deg[x] % 2 == 0)
    s.insert(make_pair(dis, x)), deg[x] = (g[x].size() + (x != rt));
  for (long long i = 0; i < g[x].size(); i++) dfs(g[x][i], dis + 1);
}
void dfs_del(int x) {
  ans.push_back(x);
  mark[x] = 1;
  for (long long i = 0; i < g[x].size(); i++)
    if (!mark[g[x][i]]) dfs_del(g[x][i]);
}
int main() {
  cin >> n;
  if (!(n % 2)) return cout << "NO", 0;
  cout << "YES" << endl;
  for (long long i = 1; i < n + 1; i++) {
    scanf("%d", &x);
    ;
    if (x == 0)
      rt = i;
    else
      g[x].push_back(i), par[i] = x;
  }
  dfs(rt, 0);
  while (s.size()) {
    int u = (*s.rbegin()).second;
    s.erase(make_pair(dist[u], u));
    dfs_del(u);
    if (u != rt) update(par[u]);
  }
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  ;
}