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 = 2e5 + 10;
const long long INF = 1e12;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int M = 1e9 + 9;
int de[N];
vector<int> G[N];
int num[N], vis[N];
void cot(int u, int p) {
  num[u] = 1;
  for (int v : G[u])
    if (v != p) {
      cot(v, u);
      num[u] += num[v];
    }
}
void dfs(int u, int p) {
  for (int v : G[u])
    if (v != p && !(num[v] & 1)) dfs(v, u);
  printf("%d ", u);
  for (int v : G[u])
    if (v != p && (num[v] & 1)) dfs(v, u);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1, j; i <= n; i++) {
    scanf("%d", &j);
    if (j) {
      de[i] ^= 1;
      de[j] ^= 1;
      G[i].push_back(j);
      G[j].push_back(i);
    }
  }
  bool f = n & 1;
  printf("%s\n", f ? "YES" : "NO");
  if (f) {
    cot(1, 0);
    dfs(1, 0);
  }
  return 0;
}