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 n;
int ans[200000 + 5], root, vis[200000 + 5], del[200000 + 5];
int degree[200000 + 5], parent[200000 + 5];
vector<int> edge[200000 + 5];
int top;
void Del(int node) {
  ans[++top] = node;
  del[node] = 1;
  for (int i = 0; i < edge[node].size(); i++) {
    int v = edge[node][i];
    if (del[v]) continue;
    degree[v] -= 1;
    if (degree[v] % 2 == 0 && parent[node] != v) Del(v);
  }
}
void dfs(int node) {
  vis[node] = 1;
  for (int i = 0; i < edge[node].size(); i++) {
    int v = edge[node][i];
    if (vis[v]) continue;
    parent[v] = node;
    dfs(v);
  }
  if (degree[node] % 2 == 0) Del(node);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int v;
    cin >> v;
    if (v == 0)
      root = i;
    else {
      edge[i].push_back(v);
      edge[v].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) degree[i] = edge[i].size();
  dfs(root);
  if (top == n) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << ans[i] << endl;
  } else
    cout << "NO";
  return 0;
}