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 nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
string nextString() {
  string s;
  cin >> s;
  return s;
}
const int MAXN = 2e5;
vector<int> G[MAXN + 10];
int num[MAXN + 10];
int f[MAXN + 10];
vector<int> order;
void dfs(int v) {
  for (auto u : G[v]) {
    dfs(u);
  }
  order.push_back(v);
}
int main() {
  int n = nextInt();
  int root = -1;
  for (int i = (1); i <= (n); ++i) {
    int x = nextInt();
    if (x == 0)
      root = i;
    else
      num[i]++, num[x]++;
    G[x].push_back(i);
    f[i] = x;
  }
  dfs(root);
  vector<int> ans0;
  vector<int> ans1;
  for (auto i : order) {
    if (num[i] & 1) {
      if (i == root) {
        cout << "NO\n";
        return 0;
      }
      ans1.push_back(i);
    } else {
      num[f[i]]--;
      ans0.push_back(i);
    }
  }
  reverse(begin(ans1), end(ans1));
  ans0.insert(ans0.end(), begin(ans1), end(ans1));
  cout << "YES\n";
  for (auto i : ans0) {
    cout << i << endl;
  }
}