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> edges[300000];
int n, m, x, y, z, p, q, r;
int sbtree[300000], isTaken[300000], degree[300000], bdchld[300000];
vector<int> anss;
void dfs1(int pos) {
  sbtree[pos]++;
  for (int to : edges[pos]) {
    dfs1(to);
    sbtree[pos] += sbtree[to];
  }
}
void dfs(int pos) {
  bool flg = true;
  for (int to : edges[pos]) {
    if (sbtree[to] % 2 == 0) {
      dfs(to);
    }
  }
  anss.push_back(pos);
  for (int to : edges[pos]) {
    if (sbtree[to] % 2) {
      dfs(to);
    }
  }
}
int main() {
  int root = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    edges[x].push_back(i);
    degree[x]++;
    degree[i]++;
    if (x == 0) root = i;
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  dfs1(root);
  dfs(root);
  for (int xx : anss) printf("%d\n", xx);
  return 0;
}
