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<int> adj[200000 + 5], stk;
bool mark[200000 + 5];
int root;
void DFSD(int s) {
  if (mark[s]) {
    return;
  }
  mark[s] = true;
  stk.push_back(s);
  for (int i = 0; i < adj[s].size(); i++) {
    DFSD(adj[s][i]);
  }
}
void DFS(int s) {
  int deg = 0;
  if (s != root) deg++;
  for (int i = 0; i < adj[s].size(); i++) {
    DFS(adj[s][i]);
    if (!mark[adj[s][i]]) {
      deg++;
    }
  }
  if (deg % 2 == 0) {
    DFSD(s);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    adj[tmp].push_back(i);
    if (tmp == 0) {
      root = i;
    }
  }
  DFS(root);
  if (stk.size() < n) {
    cout << "NO" << endl;
  } else {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      printf("%d\n", stk[i]);
    }
  }
  return 0;
}