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>
const long long MOD = 1e9 + 9;
const int MAX_N = 2 * 1e5 + 100;
const long long INF = 1e18;
const long long delta = 10957;
using namespace std;
set<int> g[MAX_N];
vector<int> moves;
set<int> todo[MAX_N];
int mark[MAX_N];
void dfs(int v, int p) {
  if (g[v].size() <= 1) return;
  set<int> nows;
  if (!mark[v]) {
    set<int> st = g[v];
    for (auto x : st) {
      if (x != p) dfs(x, v);
    }
    int d = g[v].size();
    if (d % 2) {
      mark[v]++;
      todo[v] = nows;
      return;
    }
  } else
    nows = todo[v];
  moves.push_back(v);
  for (auto x : g[v]) {
    g[x].erase(v);
    if (x != p and g[x].size() > 0) dfs(x, v);
  }
  g[v].clear();
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie();
  cin.tie();
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p;
    if (p) {
      g[i].insert(p);
      g[p].insert(i);
    }
  }
  if (n == 1) return cout << "YES\n1", 0;
  if (n % 2 == 0) return cout << "NO", 0;
  int v;
  for (int i = 1; i <= n; ++i) {
    if (g[i].size() % 2 == 0) {
      v = i;
      break;
    }
  }
  dfs(v, 0);
  cout << "YES\n";
  set<int> st;
  for (int i = 1; i <= n; ++i) st.insert(i);
  for (auto x : moves) {
    st.erase(x);
    cout << x << endl;
  }
  for (auto x : st) cout << x << endl;
  return 0;
}