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 + 5;
int n, p[maxn], r;
vector<int> ad[maxn];
vector<int> re;
bool ok[maxn];
void dfs2(int u) {
  ok[u] = true;
  re.push_back(u);
  for (auto v : ad[u])
    if (!ok[v]) {
      dfs2(v);
    }
}
bool dfs(int u) {
  int x = 0;
  for (auto v : ad[u]) {
    if (dfs(v)) x++;
  }
  int y = x + ad[u].size() + 1;
  if (u == r) y--;
  if (y % 2)
    return false;
  else {
    dfs2(u);
    return true;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n + 1; ++i) {
    cin >> p[i];
    if (p[i]) {
      ad[p[i]].push_back(i);
    } else
      r = i;
  }
  dfs(r);
  if (re.size() == n) {
    cout << "YES\n";
    for (auto x : re) cout << x << endl;
  } else
    cout << "NO";
  return 0;
}
