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 maxn = 2e5 + 20;
vector<int> adj[maxn], a[maxn], b[maxn], c[maxn], ans;
bool dp[maxn], pd[maxn];
void dfs(int v, int p = -1) {
  for (auto u : adj[v])
    if (u != p) {
      dfs(u, v);
      if (dp[u] && pd[u]) a[v].push_back(u);
      if (dp[u] && !pd[u]) b[v].push_back(u);
      if (!dp[u] && pd[u]) c[v].push_back(u);
      if (!dp[u] && !pd[u]) {
        cout << "NO" << endl;
        exit(0);
      }
    }
  if (a[v].empty()) {
    if ((int)c[v].size() % 2 != 0) dp[v] = 1;
    if ((int)c[v].size() % 2 == 0) pd[v] = 1;
  } else
    dp[v] = pd[v] = 1;
}
void print(int v, bool has, int p = -1) {
  if (a[v].empty()) {
    for (auto u : b[v]) print(u, 1, v);
    ans.push_back(v);
    for (auto u : c[v]) print(u, 0, v);
  } else {
    int k = adj[v].size();
    if (!has && p != -1) k--;
    if ((int)c[v].size() % 2 == 0) {
      for (auto u : a[v]) print(u, 1, v);
      for (auto u : b[v]) print(u, 1, v);
      ans.push_back(v);
      for (auto u : c[v]) print(u, 0, v);
    } else {
      int node = a[v].back();
      a[v].pop_back();
      for (auto u : a[v]) print(u, 1, v);
      for (auto u : b[v]) print(u, 1, v);
      ans.push_back(v);
      print(node, 0, v);
      for (auto u : c[v]) print(u, 0, v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    p--;
    if (p != -1) adj[p].push_back(i), adj[i].push_back(p);
  }
  dfs(0);
  if (!pd[0]) return cout << "NO" << endl, 0;
  cout << "YES" << endl;
  print(0, 0);
  for (auto v : ans) cout << v + 1 << endl;
}