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, rt;
vector<int> Adj[200005];
bool dp[200005];
void dfs0(int u) {
  for (int k = 0; k < Adj[u].size(); k++) {
    int v = Adj[u][k];
    dfs0(v);
    if (dp[v] == 0) dp[u] = !dp[u];
  }
}
vector<int> od;
void dfs1(int u) {
  for (int k = 0; k < Adj[u].size(); k++) {
    int v = Adj[u][k];
    if (dp[v] == 1) dfs1(v);
  }
  od.push_back(u);
  for (int k = 0; k < Adj[u].size(); k++) {
    int v = Adj[u][k];
    if (dp[v] == 0) dfs1(v);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N;
  int u, v;
  for (int i = 1; i <= N; i++) {
    cin >> v;
    if (v == 0)
      rt = i;
    else
      Adj[v].push_back(i);
  }
  dfs0(rt);
  if (dp[rt]) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  dfs1(rt);
  for (int k = 0; k < od.size(); k++) cout << od[k] << endl;
  return 0;
}