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 long long mod = 1e9 + 7;
long long powmod(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    n >>= 1;
  }
  return ans;
}
long long inv(long long a) { return powmod(a, mod - 2); }
vector<vector<long long> > g;
vector<long long> sz;
vector<long long> par;
void dfs(long long u, long long p) {
  for (auto v : g[u]) {
    if (v != p && sz[v] % 2 == 0) dfs(v, u);
  }
  cout << u + 1 << "\n";
  for (auto v : g[u]) {
    if (v != p && sz[v] % 2 == 1) dfs(v, u);
  }
}
void find_size(long long u, long long p) {
  sz[u] = 1;
  for (auto v : g[u]) {
    if (v != p) {
      find_size(v, u);
      sz[u] += sz[v];
    }
  }
}
void solve() {
  long long n;
  cin >> n;
  g.resize(n);
  par.resize(n);
  sz.resize(n);
  if (n % 2 == 0) {
    cout << "NO\n";
    return;
  }
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    x--;
    if (x >= 0) {
      g[i].push_back(x);
      g[x].push_back(i);
    }
  }
  cout << "YES\n";
  find_size(0, -1);
  dfs(0, -1);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout << setprecision(20) << fixed;
  long long t = 1;
  while (t--) {
    solve();
  }
}