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 long long N = 2e5 + 5;
vector<long long> g[N];
bool vis[N];
long long child[N];
vector<long long> ans;
void dfs(long long node) {
  vis[node] = true;
  child[node] = 1;
  for (long long x : g[node]) {
    if (!vis[x]) {
      dfs(x);
      child[node] += child[x];
    }
  }
}
void dfs2(long long node) {
  vis[node] = true;
  for (long long x : g[node]) {
    if (!vis[x] && child[x] % 2 == 0) {
      dfs2(x);
    }
  }
  ans.push_back(node);
  for (long long x : g[node]) {
    if (!vis[x] && child[x] % 2 == 1) {
      dfs2(x);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  set<long long> even, odd;
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long p;
    cin >> p;
    if (p != 0) {
      g[p].push_back(i);
      g[i].push_back(p);
    }
  }
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  dfs(1);
  memset(vis, false, sizeof(vis));
  dfs2(1);
  cout << "YES\n";
  for (long long x : ans) cout << x << "\n";
}