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;
const int mxn = 2e5 + 20;
int vis[mxn], degree[mxn], fat[mxn];
vector<int> adj[mxn];
vector<int> ans;
stack<int> rev_topo;
void topo_dfs(int p, int u) {
  rev_topo.push(u);
  fat[u] = p;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) continue;
    topo_dfs(u, v);
  }
}
void real_dfs(int u) {
  ans.push_back(u);
  vis[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    --degree[v];
    if (v == fat[u]) continue;
    if (vis[v]) continue;
    if (degree[v] % 2 == 0) real_dfs(v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  memset(vis, 0, sizeof(vis));
  memset(degree, 0, sizeof(degree));
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x) {
      adj[i].push_back(x);
      adj[x].push_back(i);
      degree[i]++;
      degree[x]++;
    }
  }
  topo_dfs(0, 1);
  while (!rev_topo.empty()) {
    int u = rev_topo.top();
    rev_topo.pop();
    if (degree[u] % 2 == 0) real_dfs(u);
  }
  if (ans.size() == n) {
    cout << "YES\n";
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << endl;
    }
  } else {
    cout << "NO" << endl;
  }
  return 0;
}