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> son[200050];
bool deleted[200500];
int seq[200050];
int dcnt, root;
void dfs2(int u) {
  if (deleted[u]) return;
  deleted[u] = 1;
  seq[++dcnt] = u;
  for (int i = 0; i < son[u].size(); ++i) dfs2(son[u][i]);
}
void dfs(int u) {
  int d = 0;
  if (u != root) d++;
  for (int i = 0; i < son[u].size(); ++i) {
    dfs(son[u][i]);
    if (!deleted[son[u][i]]) d++;
  }
  if (d % 2 == 0) dfs2(u);
  return;
}
int main() {
  int n, fa;
  scanf("%d", &n);
  for (int j = 1; j <= n; ++j) {
    scanf("%d", &fa);
    if (!fa)
      root = j;
    else
      son[fa].push_back(j);
  }
  dfs(root);
  if (dcnt != n)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= dcnt; ++i) printf("%d\n", seq[i]);
  }
}
