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.

<image>
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxn = 2e5 + 5;
vector<int> a[maxn];
int dp[maxn];
int vis[maxn];
vector<int> ans;
void getans(int t, int pre) {
  if (vis[t]) return;
  vis[t] = 1;
  ans.push_back(t);
  for (auto it : a[t]) {
    if (it == pre) continue;
    getans(it, t);
  }
}
int dfs(int t, int pre) {
  int flag = 0;
  if (pre != -1 && a[t].size() == 1) {
    dp[t] = 1;
    return 1;
  }
  int cnt = 0;
  for (auto it : a[t]) {
    if (it == pre) continue;
    dfs(it, t);
    if (dp[it] == 1) cnt++;
  }
  if (pre != -1) cnt++;
  if (cnt & 1) {
    dp[t] = 1;
    return 1;
  } else {
    dp[t] = 0;
    getans(t, pre);
    return 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int i, j, k, m, n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> k;
    if (k == 0) continue;
    a[i].push_back(k);
    a[k].push_back(i);
  }
  int t = dfs(1, -1);
  getans(1, -1);
  if (t == 1) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (auto it : ans) cout << it << " ";
  return 0;
}
