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;
const long long mod = 1e9 + 7;
const double PI = acos(-1);
int n, i, j, k, t;
vector<int> adj[200002], ans;
bool deleted[200002];
int sz[200002];
void dfs(int v, int p) {
  for (int x : adj[v]) {
    if (x == p) continue;
    if (sz[x] % 2 == 0) {
      dfs(x, v);
      deleted[x] = true;
    }
  }
  ans.push_back(v);
  for (int x : adj[v]) {
    if (x == p || deleted[x]) continue;
    dfs(x, v);
  }
}
void getsz(int v, int p) {
  sz[v] = 1;
  for (int x : adj[v]) {
    if (x == p) continue;
    getsz(x, v);
    sz[v] += sz[x];
  }
}
int main() {
  memset(deleted, false, sizeof deleted);
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &t);
    if (t) {
      adj[t].push_back(i);
      adj[i].push_back(t);
    }
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  getsz(1, 1);
  dfs(1, 1);
  puts("YES");
  for (int x : ans) printf("%d\n", x);
  return 0;
}
