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;
stack<int> s;
int deg[200005];
int vis[200005];
int p[200005];
vector<int> ne[200005];
vector<int> ans;
void dfs(int u) {
  s.push(u);
  for (int i = 0; i < ne[u].size(); i++) {
    dfs(ne[u][i]);
  }
}
void dfs2(int u) {
  vis[u] = 1;
  ans.push_back(u);
  for (int i = 0; i < ne[u].size(); i++) {
    int v = ne[u][i];
    deg[v]--;
    if (!vis[v] && deg[v] % 2 == 0) {
      dfs2(v);
    }
  }
  deg[p[u]]--;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    deg[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    vis[i] = 0;
    cin >> p[i];
    ne[p[i]].push_back(i);
    if (p[i]) {
      deg[i]++;
      deg[p[i]]++;
    }
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  int root = ne[0][0];
  dfs(root);
  while (!s.empty()) {
    int u = s.top();
    s.pop();
    if (!vis[u] && deg[u] % 2 == 0) {
      dfs2(u);
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}