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;
using ll = long long;
const int maxn = 2e5, mod = 1e9 + 9;
int n, par[maxn];
vector<int> g[maxn], ord;
void dfs1(int v) {
  par[v] ^= 1;
  ord.push_back(v);
  for (int to : g[v]) {
    if (par[to] & 1) dfs1(to);
  }
}
void dfs(int v, int p) {
  for (int to : g[v]) {
    if (to != p) dfs(to, v);
  }
  if (!par[v]) {
    ord.push_back(v);
    for (int to : g[v]) {
      if (to != p && par[to]) dfs1(to);
    }
    if (p != -1) par[p] ^= 1;
  }
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int v;
    cin >> v;
    v--;
    if (v != -1) {
      g[i].push_back(v);
      g[v].push_back(i);
    }
  }
  for (int i = 0; i < n; i++) par[i] = ((int)((g[i]).size()) & 1);
  dfs(0, -1);
  if ((int)((ord).size()) != n)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int v : ord) cout << v + 1 << '\n';
  }
}