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;
vector<int> g[(int)3e5 + 500];
int del[(int)3e5 + 500];
void dfs1(int u) {
  int cnt = 0;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    dfs1(v);
    if (!del[v]) cnt++;
  }
  if (cnt % 2 == 1) {
    cout << u << endl;
    del[u] = 1;
  }
}
void dfs2(int u) {
  if (!del[u]) cout << u << endl;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    dfs2(v);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    int rt, u;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
      del[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &u);
      if (u == 0)
        rt = i;
      else
        g[u].push_back(i);
    }
    if (n % 2 == 0)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      dfs1(rt);
      dfs2(rt);
    }
  }
  return 0;
}