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 = 1e9 + 7;
vector<vector<int> > g;
vector<int> s, d, ans;
deque<int> cur;
bool nope = false;
void dfs(const int pre, const int me) {
  for (const int son : g[me]) {
    dfs(me, son);
    if (nope) return;
  }
  if (s[me] % 2 == 0 && pre != -1) {
    d[pre] += d[me] + 1;
    ++s[pre];
    cur.push_front(me + 1);
  } else if ((s[me] % 2 == 0 && pre == -1) || (s[me] % 2 == 1 && pre != -1)) {
    ans.push_back(me + 1);
    while (d[me] > 0) {
      --d[me];
      ans.push_back(cur.front());
      cur.pop_front();
    }
  } else
    nope = true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a, r;
  cin >> n;
  g.resize(n);
  s.resize(n, 0);
  d.resize(n, 0);
  for (int i = 0; i < (int)n; ++i) {
    cin >> a;
    --a;
    if (a == -1)
      r = i;
    else
      g[a].push_back(i);
  }
  dfs(-1, r);
  if (nope)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (int i = 0; i < (int)(int)(ans).size(); ++i) cout << ans[i] << endl;
  }
  return 0;
}