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> v[200005];
long long sz[200005];
long long vis[200005];
void pnt(long long n) {
  long long i, j;
  vis[n] = 1;
  long long len = v[n].size();
  for (i = 0; i < len; i++) {
    if (!vis[v[n][i]]) pnt(v[n][i]);
    sz[n] += sz[v[n][i]];
  }
}
void dfs(long long n) {
  long long i, j;
  long long len = v[n].size();
  for (i = 0; i < len; i++) {
    if (sz[v[n][i]] % 2 == 0) dfs(v[n][i]);
  }
  printf("%lld\n", n);
  for (i = 0; i < len; i++) {
    if (sz[v[n][i]] % 2 == 1) dfs(v[n][i]);
  }
}
int main() {
  long long i, j, n, m, t, src;
  for (i = 1; i <= 200000; i++) sz[i] = 1;
  scanf("%lld", &t);
  if (t % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (i = 1; i <= t; i++) {
    scanf("%lld", &n);
    if (n == 0)
      src = i;
    else
      v[n].push_back(i);
  }
  pnt(src);
  dfs(src);
}