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 int N = 2e5 + 7;
vector<int> g[N];
int sz[N];
bool vis[N];
int lvl[N];
void dfs(int node, int par) {
  for (int nxt : g[node]) {
    if (nxt == par) continue;
    lvl[nxt] = lvl[node] + 1;
    dfs(nxt, node);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int root;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (!x) {
      root = i;
      continue;
    }
    g[x].push_back(i);
    g[i].push_back(x);
  }
  memset(sz, 0, sizeof(sz));
  memset(vis, 0, sizeof(vis));
  memset(lvl, 0, sizeof(lvl));
  dfs(root, 0);
  set<pair<int, int> > ss;
  for (int i = 1; i <= n; i++) {
    sz[i] = g[i].size();
    if (sz[i] % 2 == 0) {
      ss.insert(make_pair(lvl[i], i));
    }
  }
  vector<int> vec;
  while (ss.size()) {
    pair<int, int> p = *ss.rbegin();
    int node = p.second;
    vec.push_back(node);
    vis[node] = true;
    ss.erase(prev(ss.end()));
    for (int nxt : g[node]) {
      if (vis[nxt]) continue;
      sz[nxt]--;
      if (sz[nxt] & 1) {
        ss.erase(make_pair(lvl[nxt], nxt));
      } else
        ss.insert(make_pair(lvl[nxt], nxt));
    }
  }
  if (vec.size() != n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int x : vec) {
    cout << x << endl;
  }
  return 0;
}