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;
vector<long long int> v[200005];
long long int degree[200005], vis2[200005];
deque<long long int> dq2;
void DFS2(long long int x) {
  if (vis2[x]) {
    return;
  }
  vis2[x] = 1;
  long long int i;
  dq2.push_back(x);
  for (i = 0; i < v[x].size(); i++) {
    DFS2(v[x][i]);
  }
}
void DFS1(long long int x, long long int par) {
  long long int i;
  for (i = 0; i < v[x].size(); i++) {
    DFS1(v[x][i], x);
  }
  if (!(degree[x] % 2)) {
    DFS2(x);
    degree[par]--;
  }
}
int main(void) {
  long long int i, n, x, size = 0, lol;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    if (x != 0) {
      v[x].push_back(i);
      degree[i]++;
      degree[x]++;
    } else {
      lol = i;
    }
  }
  if (!(n % 2)) {
    cout << "NO\n";
    return 0;
  }
  DFS1(lol, 200002);
  cout << "YES\n";
  while (!dq2.empty()) {
    cout << dq2.front() << "\n";
    dq2.pop_front();
  }
  return 0;
}