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;
vector<int> graph[200002];
int visited[200002], sub[200002];
int subsize(int root) {
  visited[root] = 1;
  for (int child : graph[root]) {
    if (!visited[child]) sub[root] += subsize(child);
  }
  return sub[root];
}
void destroy(int root, int par) {
  for (int child : graph[root]) {
    if (child != par && (sub[child] % 2 == 0)) destroy(child, root);
  }
  printf("%d\n", root);
  for (int child : graph[root]) {
    if (child != par && (sub[child] & 1)) destroy(child, root);
  }
}
int main() {
  for (int i = 0; i < 200002; i++) sub[i] = 1;
  int n, x;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x) {
      graph[x].push_back(i);
      graph[i].push_back(x);
    }
  }
  if (n % 2 == 0)
    printf("NO");
  else {
    subsize(1);
    printf("YES\n");
    destroy(1, 0);
  }
  return 0;
}