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;
void die(string x) {
  cout << x;
  exit(0);
}
int n;
vector<int> g[200001];
vector<int> out[200001];
int parent[200001];
bool was[200001];
vector<int> toporder;
void rec(int x) {
  if (g[x].size() == 1 && x != 1) {
    out[parent[x]].push_back(x);
    return;
  }
  for (auto s : g[x]) {
    if (s != parent[x]) {
      parent[s] = x;
      rec(s);
    }
  }
  if (x == 1) {
    if (out[1].size() % 2 == 1) die("NO");
    return;
  }
  if (out[x].size() % 2 == 0)
    out[parent[x]].push_back(x);
  else
    out[x].push_back(parent[x]);
}
void toprec(int x) {
  was[x] = true;
  for (auto s : out[x]) {
    if (!was[s]) toprec(s);
  }
  toporder.push_back(x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x == 0) continue;
    g[i].push_back(x);
    g[x].push_back(i);
  }
  parent[1] = 1;
  rec(1);
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) {
    if (!was[i]) toprec(i);
  }
  reverse(toporder.begin(), toporder.end());
  for (auto s : toporder) cout << s << '\n';
  return 0;
}