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;
struct Edge {
  int to, next;
} e[404040];
int cnt = 0, head[202020];
void add(int u, int v) {
  cnt++;
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
int d[202020], d2[202020];
int size[202020];
void dfs(int x, int fa) {
  size[x] = 1;
  for (int i = head[x]; i != 0; i = e[i].next) {
    int to = e[i].to;
    if (to != fa) {
      dfs(to, x);
      size[x] += size[to];
      if (size[to] % 2 == 0) printf("%d\n", to);
    }
  }
}
void dfs2(int x, int fa) {
  if (size[x] % 2 == 1) printf("%d\n", x);
  for (int i = head[x]; i != 0; i = e[i].next) {
    int to = e[i].to;
    if (to != fa) dfs2(to, x);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int pi;
    scanf("%d", &pi);
    if (pi != 0) {
      d[i]++;
      d[pi]++;
      add(pi, i);
      add(i, pi);
    }
  }
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  dfs(1, 0);
  dfs2(1, 0);
  return 0;
}
