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;
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void read(long long &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
int n, m;
int pre[610000], now[310000], son[610000], tot;
int siz[310000];
bool vis[310000];
void build(int a, int b) {
  pre[++tot] = now[a];
  now[a] = tot;
  son[tot] = b;
}
int d[310000], w;
int du[310000];
void del(int x, int fa) {
  vis[x] = 1;
  d[++w] = x;
  for (int p = now[x]; p; p = pre[p])
    if ((son[p] != fa) && (vis[son[p]] == 0)) del(son[p], x);
}
void dfs(int x, int fa) {
  int flag = 1;
  for (int p = now[x]; p; p = pre[p])
    if (son[p] != fa) {
      dfs(son[p], x);
      if (vis[son[p]] == 0) {
        siz[x] += siz[son[p]] + 1;
        if (siz[son[p]] % 2 == 1) flag = 0;
      } else
        du[x]--;
    }
  if ((flag) && (du[x] % 2 == 0)) del(x, fa);
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    int a;
    read(a);
    if (a == 0) continue;
    du[i]++;
    du[a]++;
    build(i, a);
    build(a, i);
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  dfs(1, 0);
  puts("YES");
  for (int i = 1; i <= w; i++) printf("%d\n", d[i]);
  return 0;
}
