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 inf = 1000000000;
long long mod = 1000000009LL;
long long mod2 = 998244353LL;
int t;
int n;
int p[200005];
int deg[200005];
int deg2[200005];
vector<int> edges[200005];
int main() {
  cin >> n;
  vector<int> ans(n, 0);
  int l = 0, r = n - 1;
  memset(deg, 0, sizeof(deg));
  memset(deg2, 0, sizeof(deg2));
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
    if (p[i] == 0) continue;
    edges[p[i]].push_back(i);
    edges[i].push_back(p[i]);
    deg[i]++;
    deg[p[i]]++;
    deg2[i]++;
    deg2[p[i]]++;
  }
  if (n == 1) {
    cout << "YES" << endl;
    cout << 1 << endl;
    return 0;
  }
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (deg[u] % 2) {
      ans[r--] = u;
    } else {
      ans[l++] = u;
      for (int i = 0; i < edges[u].size(); ++i) {
        int v = edges[u][i];
        deg[v]--;
      }
    }
    for (int i = 0; i < edges[u].size(); ++i) {
      int v = edges[u][i];
      deg2[v]--;
      if (deg2[v] == 1) {
        q.push(v);
      }
    }
  }
  if (l > r) {
    cout << "YES" << endl;
    for (int i = 0; i < ans.size(); ++i) {
      printf("%d\n", ans[i]);
    }
  } else {
    cout << "NO" << endl;
  }
  return 0;
}