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>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
const int LIM = 2e5 + 5, MOD = 1e9 + 7;
int t, n, m, k, x, y;
vector<vector<int> > v;
vector<int> deg;
vector<int> nchild;
int dfs(int i, int par) {
  int ans = 1;
  for (auto &it : v[i])
    if (it != par) ans += dfs(it, i);
  return nchild[i] = ans;
}
bool vs[LIM];
void solve(int i, int par) {
  if (vs[i]) return;
  if (deg[i] == 0) {
    cout << i + 1 << endl;
    vs[i] = 1;
    return;
  }
  int neven = 0;
  for (auto &it : v[i])
    if (!vs[it] && it != par && !(nchild[it] & 1)) neven++;
  if (neven == 0 && !(deg[i] & 1)) {
    cout << i + 1 << endl;
    vs[i] = 1;
    deg[i] = 0;
    for (auto &it : v[i]) deg[it]--;
    for (auto &it : v[i])
      if (!vs[it] && it != par) solve(it, i);
  } else {
    for (auto &it : v[i])
      if (!vs[it] && it != par) solve(it, i);
    neven = 0;
    for (auto &it : v[i])
      if (!vs[it] && it != par && !(nchild[it] & 1)) neven++;
    if (neven == 0 && !(deg[i] & 1)) {
      cout << i + 1 << endl;
      vs[i] = 1;
      deg[i] = 0;
      for (auto &it : v[i]) deg[it]--;
      for (auto &it : v[i])
        if (!vs[it] && it != par) solve(it, i);
    } else {
      return;
    }
  }
}
int main() {
  cin >> n;
  if (!(n & 1)) {
    cout << "NO" << endl;
    return 0;
  }
  v.assign(n, vector<int>(0));
  deg.assign(n, 0);
  nchild.assign(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x == 0) continue;
    x--;
    deg[x]++;
    deg[i]++;
    v[x].push_back(i);
    v[i].push_back(x);
  }
  dfs(0, -1);
  cout << "YES\n";
  solve(0, -1);
  return 0;
}
