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>
#pragma warning(disable : 4996)
using namespace std;
vector<set<int>> graph;
vector<bool> seen;
vector<bool> removed;
vector<int> removedOrder;
void dfs(int from) {
  seen[from] = true;
  set<int> edges = graph[from];
  for (auto to : edges) {
    if (!seen[to]) dfs(to);
  }
  const int deg = graph[from].size();
  if (deg % 2 == 0 && !removed[from]) {
    removed[from] = true;
    removedOrder.push_back(from);
    for (auto to : graph[from]) graph[to].erase(from);
    graph[from].clear();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  graph.assign(n, set<int>());
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p;
    if (p) {
      --p;
      graph[i].insert(p);
      graph[p].insert(i);
    }
  }
  removed.assign(n, false);
  for (int i = 0; i < n; ++i) {
    if (!removed[i]) {
      seen.assign(n, false);
      dfs(i);
    }
  }
  if (removedOrder.size() == n) {
    cout << "YES\n";
    for (auto v : removedOrder) cout << v + 1 << '\n';
  } else {
    cout << "NO\n";
  }
  return 0;
}