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 NMAX = 200005;
int n, rad, nr[NMAX], sol[NMAX], dim;
vector<int> L[NMAX];
bitset<NMAX> viz;
inline void DFS_SOLVE(int x) {
  viz[x] = 1;
  sol[++dim] = x;
  for (auto i : L[x])
    if (!viz[i]) DFS_SOLVE(i);
}
void DFS(int node) {
  nr[node] = 1;
  for (auto i : L[node]) {
    DFS(i);
    nr[node] += nr[i];
  }
  if (!(nr[node] % 2)) DFS_SOLVE(node);
}
int main() {
  int x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (!x)
      rad = i;
    else
      L[x].push_back(i);
  }
  DFS(rad);
  DFS_SOLVE(rad);
  if (dim > n)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1; i <= dim; i++) cout << sol[i] << "\n";
  }
  return 0;
}