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.

<image>
#include <bits/stdc++.h>
using namespace std;
vector<int> ch[200005];
int sz[200005];
void getsize(int u, int pre) {
  sz[u] = 1;
  for (int i = 0; i < ch[u].size(); ++i) {
    if (ch[u][i] != pre) {
      getsize(ch[u][i], u);
      sz[u] += sz[ch[u][i]];
    }
  }
}
void dfs(int u, int pre) {
  for (int i = 0; i < ch[u].size(); i++) {
    if (ch[u][i] != pre) {
      if (sz[ch[u][i]] % 2 == 0) {
        dfs(ch[u][i], u);
      }
    }
  }
  printf("%d\n", u);
  for (int i = 0; i < ch[u].size(); i++) {
    if (ch[u][i] != pre) {
      if (sz[ch[u][i]] % 2 == 1) {
        dfs(ch[u][i], u);
      }
    }
  }
}
int main() {
  int N;
  scanf("%d", &N);
  int temp;
  int root;
  for (int i = 1; i <= N; i++) {
    scanf("%d", &temp);
    if (temp != 0) {
      ch[i].push_back(temp);
      ch[temp].push_back(i);
    } else {
      root = i;
    }
  }
  if (N % 2 == 0) {
    printf("NO\n");
  } else {
    getsize(root, -1);
    printf("YES\n");
    dfs(root, -1);
  }
  return 0;
}
