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;
long long n, i, j, k, R;
vector<long long> L, C, A, P;
vector<vector<long long>> g;
void dfs(long long u = R, long long p = 0) {
  L[u] = L[p] + 1;
  for (long long v : g[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
}
void dft(long long u) {
  if (C[u] % 2 or C[u] < 0) return;
  A.push_back(u);
  C[u] = -1;
  for (long long v : g[u]) {
    C[v]--;
    if (L[u] < L[v]) dft(v);
  }
}
bool cmp(long long i, long long j) { return L[i] > L[j]; }
int main() {
  cin >> n;
  g.resize(n + 1);
  C.assign(n + 1, 0);
  L = C;
  P.resize(n);
  for (i = 1; i <= n; i++) {
    cin >> j;
    P[i - 1] = i;
    if (j) {
      C[i]++;
      C[j]++;
      g[i].push_back(j);
      g[j].push_back(i);
    } else
      R = i;
  }
  dfs();
  sort(P.begin(), P.end(), cmp);
  for (long long i : P) dft(i);
  if (A.size() != n)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (long long i : A) cout << i << "\n";
  }
}
