Problem: 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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void Read(T &x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T>
inline bool CheckMax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool CheckMin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}
const int N = 200005;
int n, h, r, t, d[N], p[N], ans[N];
vector<int> seq, adj[N];
inline void DFS(int x) {
  seq.push_back(x);
  for (auto y : adj[x]) {
    DFS(y);
  }
}
int main() {
  Read(n), t = n + 1;
  if (!(n & 1)) {
    puts("NO");
    return 0;
  }
  puts("YES");
  for (int i = 1; i <= n; ++i) {
    Read(p[i]);
    if (p[i]) {
      adj[p[i]].push_back(i), ++d[i], ++d[p[i]];
    } else {
      r = i;
    }
  }
  DFS(r);
  reverse(seq.begin(), seq.end());
  for (auto x : seq) {
    if (d[x] & 1) {
      ans[--t] = x;
    } else {
      ans[++h] = x, --d[p[x]];
    }
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}