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 a[200010];
int deg[200010];
vector<int> e[200010];
vector<int> ans;
int d[200010], c[200010];
int dfs2(int v, int p) {
  if (deg[v] & 1) return 0;
  ans.push_back(v);
  d[v] = 1;
  for (int i = 0; i < e[v].size(); i++) {
    if (!d[e[v][i]]) {
      --deg[e[v][i]];
      if (!dfs2(e[v][i], v)) return 0;
    }
  }
  return 1;
}
int dfs(int v, int p = -1) {
  for (int i = 0; i < e[v].size(); i++) {
    if (e[v][i] != p) {
      if (dfs(e[v][i], v) == -1) return -1;
    }
  }
  if (deg[v] & 1) return 0;
  ans.push_back(v);
  d[v] = 1;
  for (int i = 0; i < e[v].size(); i++) {
    if (e[v][i] == p) {
      deg[p]--;
    } else if (!d[e[v][i]]) {
      deg[e[v][i]]--;
      if (!dfs2(e[v][i], v)) return -1;
    }
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!a[i]) continue;
    e[a[i]].push_back(i);
    e[i].push_back(a[i]);
    deg[i]++;
    deg[a[i]]++;
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  if (dfs(1) == 1) {
    puts("YES");
    for (int i = 0; i < ans.size(); i++) {
      printf("%d\n", ans[i]);
    }
  } else {
    puts("NO");
  }
  return 0;
}