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;
const long maxn = 2e5 + 100;
const long long mod = 1e9 + 9;
struct node {
  long d;
  struct node *next;
} * point[maxn];
long need[maxn], fa[maxn];
bool vis[maxn], feng[maxn];
long pr[maxn], pr_count = 0;
void dfs(long d) {
  long cc = 0;
  vis[d] = true;
  struct node *p = point[d];
  while (p) {
    if (!vis[p->d]) {
      cc++;
      dfs(p->d);
      need[d] = need[d] ^ need[p->d];
    } else
      fa[d] = p->d;
    p = p->next;
  }
}
void print(long d) {
  printf("%ld\n", d);
  struct node *p = point[d];
  while (p) {
    if (fa[d] != p->d && need[p->d] == 1) print(p->d);
    p = p->next;
  }
}
void work(long d) {
  struct node *p = point[d];
  struct node *q;
  while (p) {
    if (fa[d] != p->d) work(p->d);
    p = p->next;
  }
  if (need[d] == 0) {
    printf("%ld\n", d);
    p = point[d];
    while (p) {
      if (fa[d] != p->d && need[p->d] == 1) print(p->d);
      p = p->next;
    }
  }
}
int main() {
  struct node *p;
  long n, root, i, y;
  scanf("%ld", &n);
  for (i = 1; i <= n; i++) {
    point[i] = NULL;
    need[i] = 1;
    vis[i] = false;
  }
  for (i = 1; i <= n; i++) {
    scanf("%ld", &y);
    if (y == 0)
      root = i;
    else {
      p = (struct node *)malloc(sizeof(struct node));
      p->d = y;
      p->next = point[i];
      point[i] = p;
      p = (struct node *)malloc(sizeof(struct node));
      p->d = i;
      p->next = point[y];
      point[y] = p;
    }
  }
  need[root] = 0;
  fa[root] = 0;
  dfs(root);
  if (need[root] == 1) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  work(root);
  return 0;
}
