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;
vector<vector<int> > AdjList;
const int maxn = 2e5 + 5;
vector<int> listIdx;
int jumlahTotalNode[maxn];
int countTotalNode(int node, int parent) {
  int ans = 1;
  for (int i = 0; i < AdjList[node].size(); i++) {
    int v = AdjList[node][i];
    if (v != parent) {
      ans += countTotalNode(v, node);
    }
  }
  return jumlahTotalNode[node] = ans;
}
void dfs(int node, int parent) {
  int i, j;
  for (i = 0; i < AdjList[node].size(); i++) {
    int v = AdjList[node][i];
    if (v != parent) {
      if (jumlahTotalNode[v] % 2 == 0) {
        dfs(v, node);
      }
    }
  }
  listIdx.push_back(node);
  for (i = 0; i < AdjList[node].size(); i++) {
    int v = AdjList[node][i];
    if (v != parent) {
      if (jumlahTotalNode[v] % 2 == 1) {
        dfs(v, node);
      }
    }
  }
}
int main() {
  int n, i, j;
  scanf("%d", &n);
  AdjList.assign(n + 3, vector<int>());
  int idxRoot = 0;
  for (i = 1; i <= n; i++) {
    int parent;
    scanf("%d", &parent);
    if (parent == 0) {
      idxRoot = i;
      continue;
    }
    AdjList[parent].push_back(i);
    AdjList[i].push_back(parent);
  }
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  countTotalNode(idxRoot, 0);
  dfs(idxRoot, 0);
  printf("YES\n");
  for (i = 0; i < listIdx.size(); i++) {
    printf("%d\n", listIdx[i]);
  }
  return 0;
};
