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 maxn = 2 * 1e5 + 10;
vector<int> vec[maxn];
vector<int> ans;
stack<int> st;
int pa[maxn];
int deg[maxn];
int vis[maxn];
void dfs(int u, int pre) {
  pa[u] = pre;
  st.push(u);
  for (int i = 0; i < vec[u].size(); i++) {
    int t = vec[u][i];
    if (t == pre) continue;
    dfs(t, u);
  }
}
void dfs2(int u) {
  vis[u] = 1;
  ans.push_back(u);
  for (int i = 0; i < vec[u].size(); i++) {
    int t = vec[u][i];
    deg[t]--;
    if (t == pa[u] || vis[t]) continue;
    if (deg[t] % 2 == 0) dfs2(t);
  }
}
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i <= n; i++) vec[i].clear();
    ans.clear();
    while (!st.empty()) st.pop();
    memset(deg, 0, sizeof(deg));
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      int t;
      cin >> t;
      if (t) {
        vec[i].push_back(t);
        vec[t].push_back(i);
        deg[t]++;
        deg[i]++;
      }
    }
    dfs(1, -1);
    while (!st.empty()) {
      int t = st.top();
      st.pop();
      if (deg[t] % 2 == 0) {
        dfs2(t);
      }
    }
    if (ans.size() == n) {
      cout << "YES" << endl;
      for (int i = 0; i < ans.size(); i++) cout << ans[i] << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}