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;
vector<int> graf[200100];
pair<int, pair<int, int>> tree[1 << 19];
pair<int, pair<int, int>> st[200100];
int depth = 1 << 18, n;
void dfs() {
  int str = 1;
  st[0] = make_pair(1, make_pair(0, 1));
  while (str > 0) {
    if (st[str - 1].second.first < graf[st[str - 1].first].size()) {
      ++st[str - 1].second.first;
      if (str < 2 ||
          st[str - 2].first !=
              graf[st[str - 1].first][st[str - 1].second.first - 1]) {
        st[str] =
            make_pair(graf[st[str - 1].first][st[str - 1].second.first - 1],
                      make_pair(0, 1));
        ++str;
      }
    } else {
      if (str == 1) return;
      st[str - 2].second.second += st[str - 1].second.second;
      if (st[str - 1].second.second % 2 == 0) {
        if (str >= 2) ++tree[st[str - 2].first + depth].first;
      } else {
        ++tree[st[str - 1].first + depth].first;
      }
      --str;
    }
  }
}
void rem(int del) {
  int i;
  for (int g = 0; g < graf[del].size(); ++g) {
    i = graf[del][g] + depth;
    --tree[i].first;
    while (i > 1) {
      i >>= 1;
      if (tree[i * 2].second.first == 0)
        tree[i] = tree[2 * i + 1];
      else if (tree[i * 2 + 1].second.first == 0)
        tree[i] = tree[2 * i];
      else {
        if (tree[2 * i].first == 0)
          tree[i] = tree[2 * i];
        else
          tree[i] = tree[2 * i + 1];
      }
    }
  }
  i = del + depth;
  tree[del + depth] = make_pair(0, make_pair(0, 0));
  while (i > 1) {
    i >>= 1;
    if (tree[i * 2].second.first == 0)
      tree[i] = tree[2 * i + 1];
    else if (tree[i * 2 + 1].second.first == 0)
      tree[i] = tree[2 * i];
    else {
      if (tree[2 * i].first == 0)
        tree[i] = tree[2 * i];
      else
        tree[i] = tree[2 * i + 1];
    }
  }
}
int main() {
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO";
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int i1;
    cin >> i1;
    if (i1 != 0) {
      graf[i].push_back(i1);
      graf[i1].push_back(i);
    }
  }
  cout << "YES\n";
  dfs();
  for (int i = 0; i <= n; ++i) {
    tree[depth + i].second.first = i;
  }
  for (int i = depth - 1; i >= 1; --i) {
    if (tree[i * 2].second.first == 0)
      tree[i] = tree[2 * i + 1];
    else if (tree[i * 2 + 1].second.first == 0)
      tree[i] = tree[2 * i];
    else {
      if (tree[2 * i].first == 0)
        tree[i] = tree[2 * i];
      else
        tree[i] = tree[2 * i + 1];
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << tree[1].second.first << "\n";
    rem(tree[1].second.first);
  }
  return 0;
}