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;
int N;
vector<int> G[200000];
bool dpA[200000], dpB[200000];
void dfs(int x, int p) {
  for (int t : G[x])
    if (t != p) dfs(t, x);
  int a_only = 0, both = 0;
  for (int t : G[x])
    if (t != p) {
      bool a = dpA[t], b = dpB[t];
      if (!a && !b) return;
      if (a && !b) a_only++;
      if (a && b) both++;
    }
  if ((a_only % 2 == 0) || (both > 0)) {
    dpA[x] = true;
  }
  if ((a_only % 2 == 1) || (both > 0)) {
    dpB[x] = true;
  }
}
vector<int> perm;
void restore(int x, int p, bool par) {
  vector<int> a_only, both, b_only;
  for (int t : G[x])
    if (t != p) {
      bool a = dpA[t], b = dpB[t];
      assert(a || b);
      if (a && !b) a_only.push_back(t);
      if (a && b) both.push_back(t);
      if (!a && b) b_only.push_back(t);
    }
  for (int i = 0; i < (both.size()); i++)
    if (i != 0) restore(both[i], x, true);
  for (int v : b_only) restore(v, x, true);
  if ((a_only.size() + par) % 2 == 0) {
    if (both.size() > 0) restore(both[0], x, true);
  }
  perm.push_back(x);
  for (int v : a_only) restore(v, x, false);
  if ((a_only.size() + par) % 2 != 0) {
    assert(both.size() > 0);
    restore(both[0], x, false);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < (N); i++) {
    int p;
    cin >> p;
    p--;
    if (p != -1) G[i].push_back(p), G[p].push_back(i);
  }
  dfs(0, -1);
  if (dpA[0]) {
    cout << "YES\n";
    restore(0, -1, false);
    assert(perm.size() == N);
    for (int i = 0; i < (N); i++) cout << perm[i] + 1 << "\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}