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;
long long inf = 1ll << 30;
const int N = 2e5 + 1;
vector<long long> node[N];
long long vis[N] = {}, t[N] = {}, del[N] = {}, rt, vis2[N] = {};
map<pair<long long, long long>, bool> use;
void dfs(long long s) {
  vis[s] = 1;
  long long deg = node[s].size();
  for (long long i : node[s]) {
    if (!vis[i]) {
      dfs(i);
      if (del[i]) deg--, use[{s, i}] = 1;
    }
  }
  if (deg % 2 == 0) {
    del[s] = 1;
  } else {
    if (s == rt) {
      cout << "NO\n";
      exit(0);
    }
  }
}
void dfs2(long long s) {
  vis2[s] = 1;
  for (long long i : node[s]) {
    if (!vis2[i]) {
      if (use[{s, i}])
        t[i] = t[s] - 1;
      else
        t[i] = t[s] + 1;
      dfs2(i);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  long long p;
  for (long long i = 0; i < n; i++) {
    cin >> p;
    p--;
    if (p < 0) {
      rt = i;
      continue;
    }
    node[p].push_back(i);
    node[i].push_back(p);
  }
  dfs(rt);
  dfs2(rt);
  vector<pair<long long, long long>> v;
  for (long long i = 0; i < n; i++) v.push_back({t[i], i});
  sort(v.begin(), v.end());
  cout << "YES\n";
  for (auto i : v) cout << i.second + 1 << '\n';
}