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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 3e6;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
int n;
vector<int> adj[maxn];
int par[maxn];
bool sik[maxn];
void solve(int v) {
  cout << v << '\n';
  sik[v] = 1;
  for (auto u : adj[v])
    if (!sik[u] and u != par[v]) solve(u);
}
int dfs(int v) {
  int s = 0;
  for (auto u : adj[v])
    if (u != par[v]) par[u] = v, s += dfs(u);
  if (s % 2) solve(v);
  return (++s);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> par[i];
    if (!par[i]) continue;
    adj[i].push_back(par[i]);
    adj[par[i]].push_back(i);
  }
  memset(par, 0, sizeof par);
  ;
  if (n % 2 == 0) return (cout << "NO", 0);
  ;
  cout << "YES" << '\n';
  dfs(1);
  solve(1);
  return (0);
}