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 N = 2e5 + 5;
vector<int> edges[N];
int state[N];
void cal(int prt, int cur, vector<int> &ans) {
  ans.push_back(cur);
  state[cur] = 1;
  for (int next : edges[cur]) {
    if (next == prt) continue;
    if (state[next] == 0) cal(cur, next, ans);
  }
}
int dfs(int prt, int cur, vector<int> &ans) {
  int dsum = 0;
  for (int next : edges[cur]) {
    if (next == prt) continue;
    dsum += dfs(cur, next, ans);
  }
  if (prt != 0) dsum += 1;
  if (dsum % 2 == 0) {
    cal(prt, cur, ans);
    return 0;
  } else
    return 1;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i += 1) {
    int p;
    cin >> p;
    if (p != 0) {
      edges[i].push_back(p);
      edges[p].push_back(i);
    }
  }
  vector<int> ans;
  dfs(0, 1, ans);
  if (ans.size() != n) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (int x : ans) cout << x << endl;
  }
}