#include <bits/stdc++.h>
using namespace std;
const int N = 200002;
int dp[N];
vector<int> ans;
vector<int> g[N];
void dfs(int s, int p) {
  for (auto u : g[s]) {
    if (u != p) {
      dfs(u, s);
      dp[s] += dp[u];
    }
  }
  dp[s] = !(dp[s] % 2);
}
void dfs2(int s, int p) {
  for (auto u : g[s]) {
    if (u != p) {
      if (!dp[u]) dfs2(u, s);
    }
  }
  ans.push_back(s);
  for (auto u : g[s]) {
    if (u != p) {
      if (dp[u]) dfs2(u, s);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (!a) continue;
    g[a].push_back(i);
    g[i].push_back(a);
  }
  dfs(1, 1);
  dfs2(1, 1);
  if (!dp[1])
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) cout << ans[i] << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N = 200002;
int dp[N];
vector<int> ans;
vector<int> g[N];
void dfs(int s, int p) {
  for (auto u : g[s]) {
    if (u != p) {
      dfs(u, s);
      dp[s] += dp[u];
    }
  }
  dp[s] = !(dp[s] % 2);
}
void dfs2(int s, int p) {
  for (auto u : g[s]) {
    if (u != p) {
      if (!dp[u]) dfs2(u, s);
    }
  }
  ans.push_back(s);
  for (auto u : g[s]) {
    if (u != p) {
      if (dp[u]) dfs2(u, s);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (!a) continue;
    g[a].push_back(i);
    g[i].push_back(a);
  }
  dfs(1, 1);
  dfs2(1, 1);
  if (!dp[1])
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) cout << ans[i] << endl;
  }
  return 0;
}

```