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 N = 200002;
bool dp[N];
vector<int> Ord;
bool ok = 1;
vector<int> gr[N];
void dfs(int u, int p) {
  int deg = (int)gr[u].size() - 1;
  int solved = 0;
  for (const auto &v : gr[u]) {
    if (v == p) continue;
    dfs(v, u);
    if (dp[v]) solved++;
  }
  if (!(solved & 1)) {
    dp[u] = 1;
  }
}
void dfs2(int u, int p) {
  for (const auto &v : gr[u]) {
    if (v == p) continue;
    if (!dp[v]) {
      dfs2(v, u);
    }
  }
  Ord.push_back(u);
  for (const auto &v : gr[u]) {
    if (v == p) continue;
    if (dp[v]) {
      dfs2(v, u);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p;
    if (!p) continue;
    gr[i].push_back(p);
    gr[p].push_back(i);
  }
  dfs(1, 0);
  dfs2(1, 0);
  if (dp[1]) {
    cout << "YES\n";
    for (const auto &i : Ord) {
      cout << i << "\n";
    }
  } else {
    cout << "NO\n";
  }
  return 0;
}