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;
int sz[200001];
bool mark[200001];
set<int> tree[200001];
vector<int> destroy;
void predfs(int u, int p) {
  sz[u] = 1;
  for (int v : tree[u]) {
    if (v != p) {
      predfs(v, u);
      sz[u] += sz[v];
    }
  }
}
void dfs(int u, int p) {
  vector<int> rest;
  for (int v : tree[u]) {
    if (mark[v]) continue;
    if (v != p && sz[v] % 2 == 0) {
      dfs(v, u);
    } else if (v != p) {
      rest.push_back(v);
    }
  }
  mark[u] = true;
  destroy.push_back(u);
  for (int v : rest) {
    dfs(v, -1);
  }
}
int main() {
  int n, p;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> p;
    if (p) {
      tree[i].insert(p);
      tree[p].insert(i);
    }
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  fill(mark, mark + n + 1, false);
  predfs(1, -1);
  dfs(1, -1);
  cout << "YES" << endl;
  for (int u : destroy) {
    cout << u << endl;
  }
  return 0;
}
