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;
void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int N = 1e6;
const long long inf = 1e9 + 9;
const long double PI = acos(-1);
long long bin_pow(long long x, long long y) {
  long long res = 1, base = x;
  while (y) {
    if (y & 1) {
      res *= base;
      res %= inf;
    }
    base *= base;
    base %= inf;
    y /= 2;
  }
  return res;
}
int n;
vector<int> g[N];
int sz[N], cnt[N];
bool used[N];
queue<int> q;
void dfs(int v, int pr = -1) {
  sz[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == pr) {
      continue;
    }
    dfs(to, v);
    sz[v] += sz[to];
    if (sz[to] % 2 == 0) {
      cnt[v]++;
    }
  }
  if (v != 1 && (sz[v] & 1)) {
    cnt[v]++;
  }
}
int main() {
  boost();
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x != 0) {
      g[x].push_back(i);
      g[i].push_back(x);
    }
  }
  dfs(1);
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 0) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    ans.push_back(v);
    used[v] = 1;
    for (int i = 0; i < g[v].size(); i++) {
      int to = g[v][i];
      if (used[to]) {
        continue;
      }
      cnt[to]--;
      if (cnt[to] == 0) {
        q.push(to);
      }
    }
  }
  cout << "YES\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << "\n";
  }
}