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 N = 200001;
const long long MOD = 1e9 + 7;
int n, root, sz[N];
vector<vector<int> > g;
vector<int> ans;
void dfs1(int u, int p) {
  sz[u] = 1;
  for (int i = 0; i < g[u].size(); i++)
    if (g[u][i] != p) {
      dfs1(g[u][i], u);
      sz[u] += sz[g[u][i]];
    }
}
void dfs2(int u, int p) {
  for (int i = 0; i < g[u].size(); i++)
    if (g[u][i] != p && !(sz[g[u][i]] & 1)) dfs2(g[u][i], u);
  ans.push_back(u);
  for (int i = 0; i < g[u].size(); i++)
    if (g[u][i] != p && (sz[g[u][i]] & 1)) dfs2(g[u][i], u);
}
int main() {
  scanf("%d", &n);
  g.resize(n);
  for (int a = 0, b; a < n; a++) {
    scanf("%d", &b);
    b--;
    if (b != -1) {
      g[a].push_back(b);
      g[b].push_back(a);
    } else
      root = a;
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  dfs1(root, -1);
  dfs2(root, -1);
  puts("YES");
  for (int i = 0; i < ans.size(); i++) printf("%d\n", ans[i] + 1);
  return 0;
}