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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-6;
const long long INF = 1e9;
const long long LINF = 1e18;
const long long mod = 1e9 + 9;
const long long MAX = 2e5 + 47;
int root;
vector<int> g[MAX];
bool A[MAX], IS[MAX];
void dfs(int v, int par, bool d) {
  if (d) {
    cout << v << '\n';
    IS[v] = false;
    for (int to : g[v])
      if (to != par && IS[to]) dfs(to, v, 1);
    return;
  }
  for (int to : g[v]) {
    if (to != par && IS[to]) {
      dfs(to, v, d);
      A[v] ^= (A[to] ^ 1) && IS[to];
    }
  }
  if (A[v] ^ (v == root)) {
    cout << v << '\n';
    IS[v] = false;
    for (int to : g[v])
      if (to != par && IS[to]) dfs(to, v, 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  if (!(n & 1)) {
    cout << "NO";
    return 0;
  }
  for (int i = (1); i < (n + 1); ++i) {
    IS[i] = true;
    int p;
    cin >> p;
    if (p) {
      g[p].push_back(i);
      g[i].push_back(p);
    } else
      root = i;
  }
  cout << "YES\n";
  dfs(root, -1, 0);
  return 0;
}