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>
#pragma GCC optimize(3)
using namespace std;
int n, root, tag[300000], fa[300000];
vector<int> G[300000];
stack<int> S;
int DFS(int now) {
  int chi = G[now].size() - 1;
  int cnt = 0;
  for (int i = 0; i <= chi; ++i) {
    int to = G[now][i];
    if (to == fa[now]) continue;
    int now = DFS(to);
    cnt += now;
  }
  if (cnt & 1 && !fa[now]) {
    printf("NO\n");
    exit(0);
  }
  if (cnt & 1 || (!(cnt & 1) && !fa[now])) {
    tag[now] = 1;
    S.push(now);
    return 0;
  } else
    return 1;
}
void DFS2(int now) {
  tag[now] = 1;
  for (int i = 0; i < (int)G[now].size(); i++) {
    if (G[now][i] == fa[now] || !tag[G[now][i]]) continue;
    DFS2(G[now][i]);
  }
  printf("%d\n", now);
  for (int i = 0; i < (int)G[now].size(); i++) {
    if (G[now][i] == fa[now] || tag[G[now][i]]) continue;
    DFS2(G[now][i]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", fa + i);
    if (!fa[i]) {
      root = i;
      continue;
    }
    G[i].push_back(fa[i]);
    G[fa[i]].push_back(i);
  }
  DFS(root);
  if (!S.size()) return 0 * puts("NO");
  printf("YES\n");
  DFS2(root);
  return 0;
}
