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;
int inf = pow(10, 9);
long long int modulo = pow(10, 9) + 9;
long double eps = 1e-15;
ifstream in;
ofstream out;
vector<int> g[200100];
vector<int> taken;
vector<int>* dsu[200100];
int depth[200100];
bool J(int& a, int& b) { return depth[a] < depth[b]; }
void add(vector<int>*& a, vector<int>*& b) {
  if ((*b).size() > (*a).size()) {
    vector<int>* temp = a;
    a = b;
    b = temp;
  }
  for (int j = 0; j < (*b).size(); j++) (*a).push_back((*b)[j]);
}
bool dfs(int hd, int pr, int dep) {
  dsu[hd] = new vector<int>();
  depth[hd] = dep;
  int nece = 0;
  for (long long int i = 0; i < g[hd].size(); i++) {
    int hr = g[hd][i];
    if (hr == pr) continue;
    bool ans = dfs(hr, hd, dep + 1);
    nece += !ans;
    add(dsu[hd], dsu[hr]);
  }
  if (hd) {
    if (nece & 1) {
      sort((*dsu[hd]).begin(), (*dsu[hd]).end(), J);
      taken.push_back(hd);
      for (int j = 0; j < (*dsu[hd]).size(); j++)
        taken.push_back((*dsu[hd])[j]);
      (*dsu[hd]).clear();
      return 1;
    }
    (*dsu[hd]).push_back(hd);
    return 0;
  } else {
    if (nece & 1) return 0;
    taken.push_back(hd);
    sort((*dsu[hd]).begin(), (*dsu[hd]).end(), J);
    for (int j = 0; j < (*dsu[hd]).size(); j++) taken.push_back((*dsu[hd])[j]);
    (*dsu[hd]).clear();
    return 1;
  }
}
void deal() {
  int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    int pr;
    cin >> pr;
    --pr;
    if (pr > -1) g[pr].push_back(i), g[i].push_back(pr);
  }
  if (dfs(0, -1, 0)) {
    cout << "YES\n";
    for (int j = 0; j < taken.size(); j++) cout << taken[j] + 1 << '\n';
  } else
    cout << "NO\n";
}
int main() { deal(); }
