#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 5;
int n;
vector<int> adj[maxn];
int deg[maxn];
bool used[maxn];
int sz[maxn];
void dfsSize(int i, int p) {
  sz[i] = 1;
  for (int j : adj[i]) {
    if (j == p) continue;
    dfsSize(j, i);
    sz[i] += sz[j];
  }
}
vector<int> ans;
void dfs(int i, int p) {
  for (int j : adj[i]) {
    if (j == p) continue;
    if (!(sz[j] & 1)) {
      dfs(j, i);
    }
  }
  ans.push_back(i);
  for (int j : adj[i]) {
    if (j == p) continue;
    if (sz[j] & 1) {
      dfs(j, i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  if (!(n & 1)) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p != 0) {
      adj[i].push_back(p);
      adj[p].push_back(i);
      deg[i]++;
      deg[p]++;
    }
  }
  dfsSize(1, 1);
  dfs(1, 1);
  cout << "YES\n";
  for (int i : ans) {
    cout << i << '\n';
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 5;
int n;
vector<int> adj[maxn];
int deg[maxn];
bool used[maxn];
int sz[maxn];
void dfsSize(int i, int p) {
  sz[i] = 1;
  for (int j : adj[i]) {
    if (j == p) continue;
    dfsSize(j, i);
    sz[i] += sz[j];
  }
}
vector<int> ans;
void dfs(int i, int p) {
  for (int j : adj[i]) {
    if (j == p) continue;
    if (!(sz[j] & 1)) {
      dfs(j, i);
    }
  }
  ans.push_back(i);
  for (int j : adj[i]) {
    if (j == p) continue;
    if (sz[j] & 1) {
      dfs(j, i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  if (!(n & 1)) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p != 0) {
      adj[i].push_back(p);
      adj[p].push_back(i);
      deg[i]++;
      deg[p]++;
    }
  }
  dfsSize(1, 1);
  dfs(1, 1);
  cout << "YES\n";
  for (int i : ans) {
    cout << i << '\n';
  }
}

```