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 int m = 1e9 + 7;
int deg[200001];
int depth[200001];
vector<vector<int>> v;
void dfs(int u, int p) {
  if (p != -1) depth[u] = depth[p] + 1;
  for (int i = 0; i < v[u].size(); ++i) {
    if (v[u][i] != p) dfs(v[u][i], u);
  }
}
int main() {
  int n;
  cin >> n;
  v.resize(n);
  int root = -1;
  for (int i = 1; i <= n; ++i) {
    int a = i - 1, b;
    cin >> b;
    b--;
    if (b != -1) {
      v[a].push_back(b);
      v[b].push_back(a);
      ++deg[a];
      ++deg[b];
    } else {
      root = a;
    }
  }
  dfs(root, -1);
  set<pair<int, int>> st;
  for (int i = 0; i < n; ++i) {
    if (deg[i] % 2 == 0) st.insert({-depth[i], i});
  }
  vector<int> sol;
  while (st.size()) {
    pair<int, int> top = *st.begin();
    st.erase(top);
    if (deg[top.second] > 0)
      for (int i = 0; i < v[top.second].size(); ++i) {
        int next = v[top.second][i];
        if (deg[next] == 0) continue;
        deg[next]--;
        if (deg[next] % 2 == 0)
          st.insert({-depth[next], next});
        else
          st.erase({-depth[next], next});
      }
    deg[top.second] = 0;
    sol.push_back(top.second + 1);
  }
  if (sol.size() == n) {
    cout << "YES" << endl;
    for (int x : sol) cout << x << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}