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;
const long long infl = 1e18 + 5;
long long n, m, k, q, x, y, f, val, t, i, j;
long long ind, cnt, sz, sm, ans, mx, mn;
long long a[1000004], vis[1000004];
vector<int> g[1000004];
void dfs1(int u, int p) {
  if (vis[u]) return;
  vis[u] = 1;
  cout << u << " ";
  for (auto it : g[u]) {
    if (it != p) dfs1(it, u);
  }
}
int dfs(int u, int p) {
  int re = 0;
  if (p != -1) re = 1;
  for (auto it : g[u]) {
    if (it != p) {
      re += dfs(it, u);
    }
  }
  if (re % 2 == 0) {
    dfs1(u, p);
    return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  cin >> n;
  for (i = 1; i < n + 1; i++) {
    cin >> x;
    if (x == 0) {
      k = i;
      continue;
    }
    g[i].emplace_back(x);
    g[x].emplace_back(i);
  }
  if (n % 2 == 0)
    cout << "NO";
  else {
    cout << "YES\n";
    dfs(k, -1);
  }
  return 0;
}
