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 Nmax = 1000005;
int n, v[Nmax], degree[Nmax], solved[Nmax];
vector<int> L[Nmax], Depend[Nmax], Sol;
void deleteNode(int node) {
  solved[node] = 1;
  Sol.push_back(node);
  for (auto it : L[node]) --degree[it];
}
void clearNode(int node) {
  deleteNode(node);
  for (auto it : Depend[node]) clearNode(it);
}
void dfs(int node, int tata) {
  vector<int> Bad, Good;
  for (auto it : L[node]) {
    if (it == tata) continue;
    dfs(it, node);
    if (solved[it]) continue;
    Depend[node].push_back(it);
  }
  if (degree[node] % 2 == 0) clearNode(node);
}
int main() {
  scanf("%d", &n);
  int rad;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x == 0)
      rad = i;
    else {
      L[x].push_back(i);
      L[i].push_back(x);
      degree[i]++;
      degree[x]++;
    }
  }
  dfs(rad, 0);
  if (Sol.size() < n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (auto it : Sol) cout << it << "\n";
  return 0;
}