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 long long MOD = 1000 * 1000 * 1000 + 7;
const int N = 2e5 + 6;
int n;
vector<int> g[N], ng[N], ans;
void dest(int v) {
  ans.push_back(v);
  for (auto to : ng[v]) {
    dest(to);
  }
}
bool dfs(int v, int root = false) {
  for (auto to : g[v]) {
    if (!dfs(to)) ng[v].push_back(to);
  }
  if ((ng[v].size() + root) % 2) {
    dest(v);
    return true;
  }
  return false;
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  int root;
  for (int i = (1); i <= (n); ++i) {
    int p;
    cin >> p;
    if (!p) root = i;
    g[p].push_back(i);
  }
  dfs(root, true);
  for (int i = (0); i <= (n - 1); ++i) {
    cout << ans[i] << endl;
    assert(i < ans.size());
  }
  return 0;
}