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 INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const int N = 1 << 18;
vector<int> G[N];
int deg[N], d[N];
bool vis[N];
void dfs(int u, int fa, int dep) {
  d[u] = dep;
  for (auto& v : G[u])
    if (v != fa) dfs(v, u, dep + 1);
}
int main() {
  int n;
  cin >> n;
  int rt = 0;
  for (int i = 1; i <= n; i++) {
    static int p;
    cin >> p;
    if (p)
      G[p].push_back(i), G[i].push_back(p), ++deg[i], ++deg[p];
    else
      rt = i;
  }
  dfs(rt, 0, 0);
  priority_queue<pair<int, int> > q;
  for (int i = 1; i <= n; i++)
    if (deg[i] % 2 == 0) q.push({d[i], i});
  vector<int> ans;
  while (!q.empty()) {
    pair<int, int> tmp = q.top();
    q.pop();
    int u = tmp.second;
    if (deg[u] & 1 || vis[u]) continue;
    vis[u] = 1;
    ans.push_back(u);
    for (auto& v : G[u])
      if (--deg[v] % 2 == 0) q.push({d[v], v});
  }
  if (ans.size() != n) return cout << "NO", 0;
  cout << "YES" << endl;
  for (auto& t : ans) cout << t << endl;
  return 0;
}