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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 200005;
int tree[N];
vector<int> con[N];
int calc_parity(int s = 1, int p = 0) {
  tree[s] = 1;
  for (int v : con[s])
    if (v != p) tree[s] ^= calc_parity(v, s);
  return tree[s];
}
void dfs(int s = 1, int p = 0) {
  for (int v : con[s])
    if (v != p && tree[v] == 0) dfs(v, s);
  printf("%d\n", s);
  for (int v : con[s])
    if (v != p && tree[v] == 1) dfs(v, s);
}
int main() {
  int n;
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &(p));
    if (p) {
      con[i].push_back(p);
      con[p].push_back(i);
    }
  }
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  calc_parity();
  dfs();
}