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;
bool vowel(char che) {
  if (che == 'A' || che == 'E' || che == 'I' || che == 'O' || che == 'U' ||
      che == 'Y')
    return true;
  else
    return false;
}
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
long long a[N], b[N], c[N], d, dis, dp[N], dg[N], pd[N];
pair<long long, long long> p;
vector<long long> vt[N], vm[N], vs;
stack<long long> st;
string s, s1;
char o;
bool mark[N], mark1[N], h;
long long n1 = 0;
void dfs(long long v, long long k, long long f) {
  pd[v] = k;
  vm[f].push_back(v);
  n1 = max(n1, f);
  f += 1;
  for (int i = 0; i < vt[v].size(); i++) {
    if (vt[v][i] != k) dfs(vt[v][i], v, f);
  }
}
void dst(long long v, long long k) {
  mark1[v] = true;
  dg[k]--;
  vs.push_back(v);
  for (int i = 0; i < vt[v].size(); i++) {
    if (vt[v][i] != k && !mark1[vt[v][i]]) dst(vt[v][i], v);
  }
}
int main() {
  long long t = 0, n, m = 0, k = 0, f = 0, sum = 0, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> m;
    if (m) {
      vt[i].push_back(m);
      vt[m].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    dg[i] = vt[i].size();
  }
  if (n % 2 == 0)
    cout << "NO\n";
  else {
    cout << "YES\n";
    dfs(1, 0, 0);
    for (int i = n1; i >= 0; i--) {
      for (int j = 0; j < vm[i].size(); j++) {
        m = vm[i][j];
        if (dg[m] % 2 == 0) dst(m, pd[m]);
      }
    }
    for (auto x : vs) {
      cout << x << endl;
    }
  }
}