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 N = 222222;
int n, p[N], f[N];
vector<int> v[N];
void dfs0(int u) {
  int i = v[u].size();
  while (i--) {
    dfs0(v[u][i]);
    f[u] ^= f[v[u][i]];
  }
  f[u] ^= 1;
}
void dfs(int u) {
  int i = v[u].size(), x;
  while (i--) {
    x = v[u][i];
    if (!f[x]) dfs(x);
  }
  cout << u << endl;
  i = v[u].size();
  while (i--) {
    x = v[u][i];
    if (f[x]) dfs(x);
  }
}
int main() {
  int i, rt;
  scanf("%d", &n);
  for (i = 1; i <= n; i = i + 1) {
    scanf("%d", p + i);
    if (p[i])
      v[p[i]].push_back(i);
    else
      rt = i;
  }
  dfs0(rt);
  if (!f[rt])
    cout << "NO\n";
  else {
    cout << "YES\n";
    dfs(rt);
  }
  return 0;
}