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 MAX = 2e5 + 5;
vector<int> adj[MAX];
int degree[MAX], level[MAX], removed[MAX];
int n;
void dfs(int node, int father) {
  for (auto &each : adj[node]) {
    if (each != father) {
      level[each] = level[node] + 1;
      dfs(each, node);
    }
  }
}
int main() {
  scanf("%d", &n);
  int root = -1;
  for (int i = int(1); i < int(n + 1); i++) {
    int p;
    scanf("%d", &p);
    if (p != 0) {
      adj[i].push_back(p);
      adj[p].push_back(i);
      degree[i]++;
      degree[p]++;
    } else {
      root = i;
    }
  }
  dfs(root, -1);
  set<pair<int, int> > pq;
  for (int i = int(1); i < int(n + 1); i++) {
    if (degree[i] % 2 == 0) {
      pq.emplace(level[i], i);
    }
  }
  vector<int> order;
  while (!pq.empty()) {
    pair<int, int> top = *pq.rbegin();
    pq.erase(prev(pq.end()));
    int node = top.second;
    for (auto &each : adj[node]) {
      if (removed[each]) {
        continue;
      }
      degree[each]--;
      if (degree[each] % 2 == 1) {
        pq.erase(pq.find(pair<int, int>(level[each], each)));
      } else {
        pq.emplace(level[each], each);
      }
    }
    order.push_back(node);
    removed[node] = 1;
  }
  if ((int)order.size() == n) {
    puts("YES");
    for (auto &each : order) {
      printf("%d\n", each);
    }
  } else {
    puts("NO");
  }
  return 0;
}