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;
const int maxn = 2e5 + 11;
int n, par[maxn], siz[maxn], deg[maxn], odd[maxn];
bool mark[maxn], seen[maxn];
vector<int> g[maxn];
vector<int> q;
void dfs(int u) {
  siz[u] = 1;
  bool ok = 1;
  for (auto x : g[u]) {
    if (x != par[u]) {
      par[x] = u;
      dfs(x);
      siz[u] ^= siz[x];
      if (siz[x]) odd[u]++;
      ok &= siz[x] & 1;
    }
  }
  deg[u] = (int)(g[u]).size() - (u != 0);
  ok &= ((int)(g[u]).size() % 2 == 0);
  if (ok) q.push_back(u);
}
void recheck(int u) {
  int d = deg[u];
  if (u && !mark[par[u]]) d++;
  if (deg[u] == odd[u] && d % 2 == 0) q.push_back(u);
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  if (n % 2 == 0)
    return cout << "NO" << endl, 0;
  else
    cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cin >> par[i];
    par[i]--;
    if (par[i] == -1) continue;
    g[par[i]].push_back(i);
    g[i].push_back(par[i]);
  }
  memset(par, 0, sizeof par);
  dfs(0);
  q.reserve(maxn);
  while (!q.empty()) {
    int now = q.back();
    q.pop_back();
    if (mark[now]) continue;
    mark[now] = 1;
    cout << now + 1 << endl;
    if (now) deg[par[now]]--;
    for (auto x : g[now]) {
      if (!mark[x]) {
        recheck(x);
      }
    }
  }
}