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.

<image>
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
const int N = 200005;
int n, deg[N], sz[N];
vector<int> G[N];
void dfs_sz(int u, int fa) {
  ++sz[u];
  for (int to : G[u])
    if (to != fa) {
      dfs_sz(to, u);
      sz[u] += sz[to];
    }
}
vector<int> ans;
bool dfs(int u, int fa) {
  for (int to : G[u])
    if (to != fa && !(sz[to] & 1)) {
      if (!dfs(to, u)) return false;
      --deg[u];
    }
  if (deg[u] & 1) return false;
  ans.push_back(u);
  for (int to : G[u])
    if (to != fa && (sz[to] & 1)) {
      --deg[to];
      if (!dfs(to, u)) return false;
    }
  return true;
}
int main() {
  scanf("%d", &n);
  int pi;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &pi);
    if (pi != 0) {
      G[i].push_back(pi), G[pi].push_back(i);
      ++deg[i], ++deg[pi];
    }
  }
  dfs_sz(1, -1);
  if (dfs(1, -1)) {
    puts("YES");
    for (int v : ans) printf("%d\n", v);
  } else
    puts("NO");
  return 0;
}
