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 = 1 << 30;
const int N = 200100;
int n, a[N], q;
int lvl[N], deg[N];
vector<int> adj[N];
priority_queue<pair<int, int>, vector<pair<int, int> > > pq;
void dfs(int u, int par = 0, int l = 0) {
  deg[u] = adj[u].size();
  lvl[u] = l;
  for (auto v : adj[u]) {
    if (v != par) dfs(v, u, l + 1);
  }
  if (deg[u] % 2 == 0) pq.push({lvl[u], u});
}
vector<int> ANS;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x == 0) continue;
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  dfs(1);
  while (pq.size()) {
    pair<int, int> u = pq.top();
    pq.pop();
    if (deg[u.second] < 0 || deg[u.second] % 2) continue;
    ANS.push_back(u.second);
    for (auto v : adj[u.second]) {
      if (deg[v] != -1) {
        deg[v]--;
        if (deg[v] % 2 == 0) pq.push({lvl[v], v});
      }
      deg[u.second] = -1;
    }
  }
  if (ANS.size() != n)
    printf("NO\n");
  else {
    printf("YES\n");
    for (auto x : ANS) printf("%d\n", x);
  }
  return 0;
}