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 main() {
  int n;
  cin >> n;
  vector<int> a;
  vector<vector<int> > gr(n + 1, a);
  int any;
  int th;
  for (int i = 1; i <= n; i++) {
    cin >> th;
    if (th != 0) {
      gr[i].push_back(th);
      gr[th].push_back(i);
    }
  }
  vector<int> chet(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    if (gr[i].size() % 2 == 0) {
      chet[i] = 1;
      any = i;
    }
  }
  if (n % 2 == 0) {
    cout << "NO";
  } else {
    cout << "YES\n";
    queue<int> q;
    q.push(any);
    vector<int> dist(n + 1, 0);
    vector<int> bfu(n + 1, 0);
    vector<int> pred(n + 1, 0);
    vector<vector<int> > next(n + 1, a);
    bfu[any] = 1;
    pred[any] = -1;
    while (q.size() != 0) {
      int v = q.front();
      q.pop();
      for (int i = 0; i < gr[v].size(); i++) {
        if (bfu[gr[v][i]] == 0) {
          pred[gr[v][i]] = v;
          q.push(gr[v][i]);
          bfu[gr[v][i]] = 1;
          next[v].push_back(gr[v][i]);
          dist[gr[v][i]] = dist[v] + 1;
        }
      }
    }
    vector<int> ans;
    int maxd = dist[max_element(dist.begin(), dist.end()) - dist.begin()];
    vector<int> ansu(n + 1, 0);
    set<int> sss;
    vector<set<int> > ch(maxd + 1, sss);
    for (int i = 1; i <= n; i++) {
      if (chet[i] == 1) {
        ch[dist[i]].insert(i);
      }
    }
    for (int i = ch.size() - 1; i >= 0; i--) {
      for (auto j : ch[i]) {
        queue<int> qq;
        qq.push(j);
        if (i != 0) {
          if (ch[i - 1].find(pred[j]) != ch[i - 1].end()) {
            ch[i - 1].erase(pred[j]);
          } else {
            ch[i - 1].insert(pred[j]);
          }
        }
        while (qq.size() != 0) {
          int v = qq.front();
          qq.pop();
          if (ansu[v] == 0) {
            cout << v << '\n';
            ansu[v] = 1;
            for (int jj = 0; jj < next[v].size(); jj++) {
              qq.push(next[v][jj]);
            }
          }
        }
      }
    }
  }
  return 0;
}