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;
int destroy[200002], vis[200002], sub[200002], parity[200002], lvl[200002];
vector<vector<int> > g(200002);
vector<int> tame;
int dfs(int k) {
  vis[k] = 1;
  sub[k] = 1;
  for (int i = 0; i < g[k].size(); i++) {
    if (vis[g[k][i]] == 0) {
      lvl[g[k][i]] = lvl[k] + 1;
      sub[k] += dfs(g[k][i]);
      if (destroy[g[k][i]] == 1) {
        parity[k] ^= 1;
      }
    }
  }
  if (parity[k] == 0) {
    tame.push_back(k);
    destroy[k] = 1;
    return 0;
  }
  return sub[k];
}
int dfs1(int k) {
  cout << k << "\n";
  vis[k] = 1;
  for (int i = 0; i < g[k].size(); i++) {
    if (vis[g[k][i]] == 0 && lvl[g[k][i]] > lvl[k]) {
      dfs1(g[k][i]);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x != 0) {
      g[i].push_back(x);
      g[x].push_back(i);
      parity[i] ^= 1;
      parity[x] ^= 1;
    }
  }
  dfs(1);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (destroy[i]) {
      cnt += sub[i];
    }
  }
  if (cnt == n) {
    cout << "YES\n";
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < tame.size(); i++) {
      dfs1(tame[i]);
    }
  } else
    cout << "NO\n";
}