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 int maxn = 213456;
int p[maxn], deg[maxn], n;
vector<int> G[maxn], t;
void dfs(int u) {
  for (int v : G[u]) dfs(v);
  t.push_back(u);
}
int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cin >> n;
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  int r = 1;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    deg[p[i]]++;
    if (p[i])
      deg[i]++;
    else
      r = i;
    G[p[i]].push_back(i);
  }
  dfs(r);
  vector<int> ans1, ans2;
  for (int u : t) {
    if (deg[u] & 1)
      ans1.push_back(u);
    else {
      ans2.push_back(u);
      deg[p[u]]--;
    }
  }
  for (int u : ans2) printf("%d\n", u);
  reverse(ans1.begin(), ans1.end());
  for (int u : ans1) printf("%d\n", u);
  return 0;
}