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.

<image>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n;
int rt, du[maxn], fa[maxn], vis[maxn];
vector<int> Map[maxn];
void del(int u) {
  cout << u << endl;
  for (auto v : Map[u])
    if (!vis[v]) del(v);
}
void dfs(int dq) {
  for (auto x : Map[dq]) dfs(x);
  if (!(du[dq] & 1)) {
    vis[dq] = 1;
    du[fa[dq]]--;
    del(dq);
    return;
  }
}
void work() {
  if (n % 2 == 0) {
    cout << "NO" << endl;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
    }
    return;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x == 0) {
      rt = i;
      continue;
    }
    Map[x].push_back(i);
    fa[i] = x;
    ++du[x], ++du[i];
  }
  dfs(rt);
}
int main() {
  while (cin >> n) work();
  return 0;
}
