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;
vector<int> aj[200005];
vector<int> ans;
int dp[200005], mm[200005];
set<pair<int, int> > ss, ss1;
void dfs(int i, int p, int d) {
  dp[i] = d;
  for (auto it : aj[i])
    if (it != p) dfs(it, i, d + 1);
}
int main() {
  int n;
  cin >> n;
  if (n % 2 == 0) return printf("NO\n") < 0;
  if (n == 1) return printf("YES\n1\n") < 0;
  int a;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    if (!a) continue;
    aj[i + 1].push_back(a);
    aj[a].push_back(i + 1);
    mm[i + 1]++;
    mm[a]++;
  }
  dfs(1, 0, 0);
  for (int i = 1; i < n + 1; i++)
    if (mm[i] % 2 == 0) ss.insert(make_pair(-dp[i], i));
  int k;
  while (!ss.empty()) {
    k = (*ss.begin()).second;
    mm[k] = 0;
    ans.push_back(k);
    ss.erase(make_pair(-dp[k], k));
    for (auto it : aj[k]) {
      if (mm[it] == 0) continue;
      if (mm[it] % 2)
        ss.insert(make_pair(-dp[it], it));
      else
        ss.erase(make_pair(-dp[it], it));
      mm[it]--;
    }
  }
  if (ans.size() != n) return printf("NO") < 0;
  cout << "YES\n";
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << endl;
  if (ans.size() == 0) cout << 1 << endl;
}