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, ta, tb;
vector<pair<int, int> > v;
vector<int> vv[200005], vvv;
stack<int> s;
bool dfs(int a, int b) {
  bool deg = 0;
  for (int i = 0; i < vv[a].size(); i++)
    if (vv[a][i] == b)
      deg ^= 1;
    else
      deg ^= dfs(vv[a][i], a);
  if (deg) {
    s.push(a);
    return 1;
  }
  vvv.push_back(a);
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    ta = i;
    scanf("%d", &tb);
    if (tb) {
      if (ta > tb) swap(ta, tb);
      v.push_back(make_pair(ta, tb));
    }
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++)
    if (!i || v[i] != v[i - 1]) {
      vv[v[i].first].push_back(v[i].second);
      vv[v[i].second].push_back(v[i].first);
    }
  if (dfs(1, 0))
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 0; i < vvv.size(); i++) printf("%d\n", vvv[i]);
    while (!s.empty()) {
      printf("%d\n", s.top());
      s.pop();
    }
  }
  return 0;
}