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;
using ll = long long;
using ld = long double;
const int N = 200100;
vector<int> g[N];
int n, deg[N], tin[N];
int timer;
void dfs(int u = 1, int par = 0) {
  tin[u] = ++timer;
  for (auto& v : g[u])
    if (v != par) {
      dfs(v, u);
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int root;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p != 0) {
      g[p].emplace_back(i);
      g[i].emplace_back(p);
    } else
      root = i;
  }
  timer = 0;
  dfs(root);
  set<pair<int, int>> x[2];
  for (int i = 1; i <= n; i++) {
    deg[i] = g[i].size();
    x[deg[i] % 2].emplace(-tin[i], i);
  }
  if (x[0].size() % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << '\n';
  while (x[0].size() > 0) {
    int u = x[0].begin()->second;
    x[0].erase(x[0].begin());
    cout << u << '\n';
    for (auto& v : g[u]) {
      if (deg[v] == 0) continue;
      x[deg[v] % 2].erase({-tin[v], v});
      deg[v]--;
      x[deg[v] % 2].emplace(-tin[v], v);
    }
    deg[u] = 0;
  }
  return 0;
}