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 INF = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
const int maxn = 4e5;
long long n, mark[maxn], p, r, val[maxn], d[maxn];
vector<long long> g[maxn];
deque<long long> dq, ans;
void dfs(long long v, long long p) {
  dq.push_front(v);
  mark[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    if (!mark[g[v][i]]) {
      dfs(g[v][i], v);
      val[v] += val[g[v][i]];
    }
  }
  val[v]++;
  if (d[v] % 2 == 0) {
    for (int i = val[v] - 1; i >= 0; i--) ans.push_back(dq[i]);
    for (int i = 0; i < val[v]; i++) dq.pop_front();
    d[p]--;
    val[v] = 0;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p;
    if (p == 0)
      r = i;
    else {
      g[p].push_back(i);
      g[i].push_back(p);
    }
  }
  for (int i = 1; i <= n; i++) d[i] = g[i].size();
  dfs(r, 0);
  if (ans.size() != n)
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) cout << ans[i] << endl;
  }
}