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;
int done[200005], deg[200005];
vector<int> adj[200005];
void dfs(int node, int par) {
  int x = 0, y = 0;
  for (int child : adj[node]) {
    if (child == par) continue;
    dfs(child, node);
    if (done[child])
      x++;
    else
      y++;
  }
  if ((deg[node] - x) % 2 == 0) {
    done[node] = 1;
  }
}
void doit(int node, int par) {
  for (int child : adj[node]) {
    if (child == par) continue;
    if (done[child]) doit(child, node);
  }
  printf("%d\n", node);
  for (int child : adj[node]) {
    if (child == par) continue;
    if (!done[child]) doit(child, node);
  }
}
int main() {
  int n, p;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p);
    adj[p].push_back(i);
    if (p) {
      adj[i].push_back(p);
    }
  }
  for (int i = 1; i <= n; i++) {
    deg[i] = adj[i].size();
  }
  dfs(adj[0][0], 0);
  if (!done[adj[0][0]] || n % 2 == 0)
    printf("NO\n");
  else {
    printf("YES\n");
    doit(adj[0][0], 0);
  }
  return 0;
}