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 p[200200];
vector<int> g[200200];
int size[200200];
vector<int> res;
void dfs(int v) {
  size[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    dfs(g[v][i]);
    size[v] += size[g[v][i]];
  }
}
void solve(int v, bool f = 0) {
  int cnt = 0;
  for (int i = 0; i < g[v].size(); i++) {
    if (size[g[v][i]] % 2 == 0) {
      cnt++;
    }
  }
  if (size[v] % 2) {
    if (f == 1) {
      printf("NO\n");
      exit(0);
    } else {
      for (int i = 0; i < g[v].size(); i++) {
        if (size[g[v][i]] % 2 == 0) solve(g[v][i], 1);
      }
      res.push_back(v);
      for (int i = 0; i < g[v].size(); i++) {
        if (size[g[v][i]] % 2 == 1) solve(g[v][i]);
      }
    }
  } else {
    if (f == 0) {
      printf("NO\n");
      exit(0);
    } else {
      for (int i = 0; i < g[v].size(); i++) {
        if (size[g[v][i]] % 2 == 0) solve(g[v][i], 1);
      }
      res.push_back(v);
      for (int i = 0; i < g[v].size(); i++) {
        if (size[g[v][i]] % 2 == 1) solve(g[v][i]);
      }
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int v;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    if (p[i] != 0)
      g[p[i]].push_back(i);
    else
      v = i;
  }
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  dfs(v);
  solve(v);
  printf("YES\n");
  for (int i = 0; i < res.size(); i++) {
    printf("%d\n", res[i]);
  }
  return 0;
}