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;
unsigned char *f;
vector<vector<int>> subs;
int i, k;
int main() {
  char dfs_f(int x);
  void dfs_print(int x);
  scanf("%d", &n);
  if (n == 1) {
    printf("YES\n1\n");
    return 0;
  } else if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  f = (unsigned char *)malloc(n + 1);
  subs.resize(n + 1);
  for (i = 0; i < n; ++i) {
    scanf("%d", &k);
    subs[k].push_back(i + 1);
  }
  k = subs[0].front();
  dfs_f(k);
  printf("YES\n");
  dfs_print(k);
  return 0;
}
char dfs_f(int x) {
  f[x] = 0;
  if (subs[x].size() == 0) return 0;
  for (vector<int>::iterator it = subs[x].begin(); it != subs[x].end(); ++it)
    if (dfs_f(*it) == 0) ++f[x];
  f[x] &= 1;
  return f[x];
}
void dfs_print(int x) {
  for (vector<int>::iterator it = subs[x].begin(); it != subs[x].end(); ++it)
    if (f[*it]) dfs_print(*it);
  printf("%d\n", x);
  for (vector<int>::iterator it = subs[x].begin(); it != subs[x].end(); ++it)
    if (!f[*it]) dfs_print(*it);
}