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;
int n, r, p[200005], vis[200005];
vector<int> vv[200005], ans;
void ddfs(int x) {
  if (vis[x]) return;
  vis[x] = 1;
  ans.push_back(x);
  for (auto i : vv[x]) {
    ddfs(i);
  }
}
void dfs(int x) {
  int num = vis[p[x]] ? 0 : 1;
  for (auto i : vv[x]) {
    dfs(i);
    if (!vis[i]) num++;
  }
  if (num % 2 == 0) ddfs(x);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    if (p[i] == 0)
      r = i;
    else
      vv[p[i]].push_back(i);
  }
  vis[0] = 1;
  dfs(r);
  if (ans.size() != n)
    puts("NO");
  else {
    puts("YES");
    for (auto i : ans) printf("%d\n", i);
  }
  return 0;
}