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 Y {
  int h, d;
  inline bool operator<(const Y& a) const { return d < a.d; }
} dep[200005];
int n, d[200005], rt, p[200005], que[200005], f[200005], cnt;
vector<int> a[200005];
void dfs(int x, int fa) {
  dep[x].h = x, dep[x].d = dep[fa].d + 1, f[x] = fa;
  for (int i = 0; i < a[x].size(); ++i)
    if (a[x][i] != fa) dfs(a[x][i], x);
}
void dfs1(int x) {
  for (int i = 0; i < a[x].size(); ++i)
    if (p[a[x][i]] == 1 && a[x][i] != f[x]) dfs1(a[x][i]);
  que[++cnt] = x;
  for (int i = 0; i < a[x].size(); ++i)
    if (p[a[x][i]] == 0 && a[x][i] != f[x]) dfs1(a[x][i]);
}
int main(void) {
  register int i, x;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (x != 0)
      ++d[i], ++d[x], a[i].push_back(x), a[x].push_back(i);
    else
      rt = i;
  }
  dfs(rt, 0);
  sort(dep + 1, dep + n + 1);
  for (i = n; i; --i) {
    if (d[dep[i].h] & 1)
      p[dep[i].h] = 0;
    else
      p[dep[i].h] = 1, --d[f[dep[i].h]];
    ;
    if (dep[i].h == rt && p[dep[i].h] == 0) {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  dfs1(rt);
  for (i = 1; i <= n; ++i) printf("%d\n", que[i]);
  return 0;
}
