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.

<image>
#include <bits/stdc++.h>
using namespace std;
vector<int> ans;
int previ[200005];
vector<int> tree[200005];
int cnt[200005];
void countN(int curr) {
  cnt[curr] = 1;
  for (int a : tree[curr]) {
    if (a != previ[curr]) {
      previ[a] = curr;
      countN(a);
      cnt[curr] += cnt[a];
    }
  }
}
void recur(int curr) {
  for (int a : tree[curr]) {
    if (a != previ[curr]) {
      if (cnt[a] % 2 == 0) {
        recur(a);
      }
    }
  }
  ans.push_back(curr);
  for (int a : tree[curr]) {
    if (a != previ[curr]) {
      if (cnt[a] % 2 == 1) {
        recur(a);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    temp--;
    if (temp != -1) {
      tree[i].push_back(temp);
      tree[temp].push_back(i);
    }
  }
  if (n % 2 == 1) {
    cout << "YES\n";
    previ[0] = -1;
    countN(0);
    recur(0);
    for (int a : ans) {
      cout << a + 1 << "\n";
    }
  } else {
    cout << "NO\n";
  }
}
