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> g[200100];
int sz[200100], root;
bool vist[200100];
void delSubtree(int u) {
  vist[u] = true;
  printf("%d\n", u);
  for (auto v : g[u])
    if (!vist[v]) delSubtree(v);
}
void DFS(int u, int p) {
  int degree = 0;
  for (auto v : g[u]) {
    if (v == p) continue;
    DFS(v, u);
    if (!vist[v]) degree++;
  }
  if (u != root) degree++;
  if (!(degree & 1)) delSubtree(u);
}
int main() {
  int n, u;
  scanf("%d", &n);
  for (int v = 1; v <= n; v++) {
    scanf("%d", &u);
    if (u)
      g[u].push_back(v);
    else
      root = v;
  }
  if (n & 1)
    printf("YES\n"), DFS(root, 0);
  else
    printf("NO");
  return 0;
}