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;
char seen[201010];
vector<int> g[201010];
int deg[201010], lvl[201010];
int n, x;
set<pair<int, int>, greater<pair<int, int> > > par;
void dfs(int u, int p) {
  lvl[u] = lvl[p] + 1;
  for (int v : g[u]) {
    if (v != p) dfs(v, u);
  }
}
int main() {
  int u;
  cin >> n;
  int r;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x) {
      g[x].push_back(i);
      g[i].push_back(x);
      deg[x]++;
      deg[i]++;
    } else
      r = i;
  }
  dfs(r, r);
  for (int i = 1; i <= n; i++)
    if (deg[i] % 2 == 0) par.insert(make_pair(lvl[i], i));
  vector<int> ans;
  while (par.size()) {
    pair<int, int> tmp = *par.begin();
    par.erase(par.begin());
    u = tmp.second;
    seen[u] = 1;
    ans.push_back(u);
    for (int v : g[u]) {
      if (seen[v]) continue;
      deg[v]--;
      if (deg[v] % 2 == 0)
        par.insert(make_pair(lvl[v], v));
      else
        par.erase(make_pair(lvl[v], v));
    }
  }
  if (ans.size() == n) {
    cout << "YES\n";
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << "\n";
  } else
    cout << "NO\n";
  return 0;
}