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;
long long power(long long x, long long y, long long m);
long long modInverse(long long n, long long m);
long long nCr(long long n, long long r, long long m);
long long ceiling(long long x, long long y);
bool sortbyth(const tuple<long long, int, int>& a,
              const tuple<long long, int, int>& b) {
  if (get<0>(a) != get<0>(b))
    return get<0>(a) > get<0>(b);
  else
    return get<1>(a) < get<1>(b);
}
int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
void aksayushx() {
  int n;
  cin >> n;
  vector<vector<int>> adj(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    --x;
    if (x >= 0) {
      adj[i].push_back(x);
      adj[x].push_back(i);
    }
  }
  vector<int> degree(n, 0);
  for (int i = 0; i < n; i++) degree[i] = (long long)adj[i].size();
  if (n % 2 == 0) {
    cout << "NO\n";
    return;
  }
  int root = -1;
  vector<int> ans, dep(n);
  vector<bool> seen(n, false);
  function<void(int, int, int)> dfs = [&](int x, int p, int d) {
    dep[x] = d;
    for (int i : adj[x]) {
      if (i != p) dfs(i, x, d + 1);
    }
  };
  dfs(0, 0, 0);
  priority_queue<pair<int, int>> q;
  for (int i = 0; i < n; i++) q.push({dep[i], i});
  while (!q.empty()) {
    pair<int, int> x = q.top();
    q.pop();
    if (seen[x.second] || degree[x.second] % 2) continue;
    seen[x.second] = true;
    ans.push_back(x.second + 1);
    for (int u : adj[x.second]) {
      if (seen[u]) continue;
      degree[u]--;
      q.push({dep[u], u});
    }
  }
  cout << "YES\n";
  for (int u : ans) cout << u << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) aksayushx();
  return 0;
}
long long power(long long x, long long y, long long m) {
  long long res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
long long modInverse(long long n, long long m) { return power(n, m - 2, m); }
long long ceiling(long long x, long long y) { return (x + y - 1) / y; }