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;
int n, bb, first[200100], size[200100], ds[200100], ans[200100], aa;
struct Bn {
  int to, next;
} bn[200100 << 1];
vector<int> vec[200100];
queue<int> que;
inline void add(int u, int v) {
  bb++;
  bn[bb].to = v;
  bn[bb].next = first[u];
  first[u] = bb;
}
void gs(int now, int last) {
  int p, q;
  size[now] = 1;
  for (p = first[now]; p != -1; p = bn[p].next) {
    if (bn[p].to == last) continue;
    gs(bn[p].to, now);
    size[now] += size[bn[p].to];
  }
}
void dfs(int now, int last) {
  int p;
  for (p = first[now]; p != -1; p = bn[p].next) {
    if (bn[p].to == last) continue;
    if (size[bn[p].to] % 2) {
      vec[now].push_back(bn[p].to);
      ds[bn[p].to]++;
    } else {
      vec[bn[p].to].push_back(now);
      ds[now]++;
    }
    dfs(bn[p].to, now);
  }
}
int main() {
  memset(first, -1, sizeof(first));
  int i, j, p, q;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &p);
    if (p) add(p, i), add(i, p);
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  gs(1, -1);
  dfs(1, -1);
  for (i = 1; i <= n; i++) {
    if (!ds[i]) que.push(i);
  }
  for (; !que.empty();) {
    p = que.front();
    que.pop();
    printf("%d\n", p);
    for (i = 0; i < vec[p].size(); i++) {
      ds[vec[p][i]]--;
      if (!ds[vec[p][i]]) {
        que.push(vec[p][i]);
      }
    }
  }
}