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>
int B = 400;
using namespace std;
const long long inf = 1e12;
int MOD = 1e9 + 9;
const int mxn = 200010;
int deg[mxn];
bool del[mxn];
vector<int> ans;
vector<int> g[mxn];
void dell(int v, int p) {
  if (del[v] != 0) return;
  if (deg[v] % 2 != 0) return;
  del[v] = 1;
  ans.push_back(v);
  for (int u : g[v]) deg[u]--;
  for (int u : g[v]) {
    if (!del[u] && u != p) dell(u, v);
  }
}
void dfs(int v, int p) {
  for (int u : g[v]) {
    if (u != p) dfs(u, v);
  }
  if (deg[v] % 2 == 0) dell(v, p);
}
long long res = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (!a) continue;
    g[i].push_back(a);
    g[a].push_back(i);
    deg[a]++, deg[i]++;
  }
  dfs(1, -1);
  if (ans.size() == n) {
    cout << "YES" << endl;
    for (int u : ans) cout << u << endl;
  } else
    cout << "NO" << endl;
}