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;
long long inf = 2147483647;
const long long ll = 9223372036854775807, ninf = 1000000000;
const double eps = 1e-6;
const long long nll = 1000000000000000000;
vector<int> node[200005];
vector<int> num;
int deg[200005] = {};
void dfs(int x) {
  for (int i = 0; i < node[x].size(); i++) {
    dfs(node[x][i]);
  }
  num.push_back(x);
}
int A[200005] = {};
int main() {
  int a;
  while (cin >> a) {
    int r = 1;
    for (int i = 1; i <= a; i++) {
      cin >> A[i];
      deg[A[i]]++;
      if (A[i]) {
        deg[i]++;
        node[A[i]].push_back(i);
      } else {
        r = i;
      }
    }
    if (a % 2 == 0) {
      cout << "NO" << '\n';
      continue;
    }
    dfs(r);
    vector<int> ans1, ans2;
    cout << "YES" << '\n';
    for (int i = 0; i < num.size(); i++) {
      int now = num[i];
      if (deg[now] % 2 == 0) {
        ans2.push_back(now);
        deg[A[now]]--;
      } else {
        ans1.push_back(now);
      }
    }
    for (int i = 0; i < ans2.size(); i++) {
      cout << ans2[i] << '\n';
    }
    reverse(ans1.begin(), ans1.end());
    for (int i = 0; i < ans1.size(); i++) {
      cout << ans1[i] << '\n';
    }
  }
  return 0;
}
