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>
inline long long read() {
  long long x = 0;
  char ch = getchar(), w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * w;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
using namespace std;
const int N = 3e5 + 666;
vector<int> G[N];
int n, siz[N];
void dfs(int x, int F) {
  siz[x] = 1;
  for (int y : G[x])
    if (y != F) {
      dfs(y, x);
      siz[x] += siz[y];
    }
}
vector<int> ans;
void del(int x, int F) {
  for (int y : G[x])
    if (y != F) {
      if (siz[y] % 2 == 0) del(y, x);
    }
  if (x) ans.push_back(x);
  for (int y : G[x])
    if (y != F) {
      if (siz[y] % 2) del(y, x);
    }
}
int main() {
  n = read();
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int x = read();
    G[x].push_back(i);
    G[i].push_back(x);
  }
  dfs(0, 0);
  del(0, 0);
  puts("YES");
  for (int i : ans) writeln(i);
  return 0;
}
