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 int maxn = 2e5 + 10;
const int maxm = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int casn, n, m, k;
int st, pre[maxn], deg[maxn];
int to[maxn], nex[maxn], head[maxn], nume;
int dfn[maxn], cnt, ans1[maxn], ans2[maxn], cnt2;
void dfs(int now) {
  for (int i = head[now]; i; i = nex[i]) {
    if (to[i] != now) dfs(to[i]);
  }
  dfn[++cnt] = now;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> pre[i];
    deg[pre[i]]++;
    if (pre[i]) {
      deg[i]++;
      to[++nume] = i;
      nex[nume] = head[pre[i]];
      head[pre[i]] = nume;
    } else
      st = i;
  }
  dfs(st);
  cnt = 0;
  for (int i = 1; i < n; i++) {
    if (deg[dfn[i]] & 1) {
      ans2[++cnt2] = dfn[i];
    } else
      ans1[++cnt] = dfn[i], deg[pre[dfn[i]]]--;
  }
  if (deg[dfn[n]] & 1)
    cout << "NO\n";
  else {
    ans1[++cnt] = dfn[n];
    cout << "YES\n";
    for (int i = 1; i <= cnt; i++) cout << ans1[i] << '\n';
    for (int i = cnt2; i; i--) cout << ans2[i] << '\n';
  }
  return 0;
}
