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.

<image>
#include <bits/stdc++.h>
using namespace std;
vector<int> graf[200005], stek;
bool mark[200005];
int root;
void dfsd(int s) {
  if (mark[s]) return;
  mark[s] = true;
  stek.push_back(s);
  for (int i = 0; i < graf[s].size(); i++) dfsd(graf[s][i]);
}
void dfs(int s) {
  int deg = 0;
  if (s != root) deg++;
  for (int i = 0; i < graf[s].size(); i++) {
    dfs(graf[s][i]);
    if (!mark[graf[s][i]]) deg++;
  }
  if (deg % 2 == 0) dfsd(s);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    graf[t].push_back(i);
    if (t == 0) root = i;
  }
  dfs(root);
  if (stek.size() < n)
    printf("NO");
  else {
    printf("YES\n");
    for (int i = 0; i < n; i++) printf("%d\n", stek[i]);
  }
}
