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 Inf = (int)1e9;
const long long InfL = (long long)1e18;
const int MaxN = (int)2e5 + 5;
vector<int> adj[MaxN], ans;
int n, a, deg[MaxN], dead[MaxN];
void dfs2(int x, int p) {
  dead[x] = true;
  ans.push_back(x);
  for (auto u : adj[x]) --deg[u];
  for (auto u : adj[x])
    if (u != p && !dead[u]) dfs2(u, x);
}
void dfs1(int x, int p) {
  for (auto u : adj[x])
    if (u != p) dfs1(u, x);
  if (!(deg[x] & 1)) dfs2(x, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    if (!a) continue;
    adj[i].push_back(a);
    deg[i]++;
    adj[a].push_back(i);
    deg[a]++;
  }
  dfs1(1, 0);
  if (ans.size() == n) {
    cout << "YES" << '\n';
    for (int i = 0; i < n; i++) cout << ans[i] << '\n';
  } else {
    cout << "NO" << '\n';
  }
  return 0;
}