#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
bool vis[N], done[N];
vector<int> res, g[N];
void solve(int u) {
  res.push_back(u);
  done[u] = 1;
  for (int v : g[u]) {
    if (done[v] == 0) {
      solve(v);
    }
  }
}
int root;
void dfs(int u) {
  int deg = (u != root);
  for (int v : g[u]) {
    dfs(v);
    deg += (!done[v]);
  }
  if (deg % 2 == 0) {
    solve(u);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p) {
      g[p].push_back(i);
    } else {
      root = i;
    }
  }
  if (n % 2 == 0) {
    cout << "NO"
         << "\n";
  } else {
    dfs(root);
    cout << "YES"
         << "\n";
    for (int x : res) {
      cout << x << " ";
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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 = 1e6 + 100;
bool vis[N], done[N];
vector<int> res, g[N];
void solve(int u) {
  res.push_back(u);
  done[u] = 1;
  for (int v : g[u]) {
    if (done[v] == 0) {
      solve(v);
    }
  }
}
int root;
void dfs(int u) {
  int deg = (u != root);
  for (int v : g[u]) {
    dfs(v);
    deg += (!done[v]);
  }
  if (deg % 2 == 0) {
    solve(u);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p) {
      g[p].push_back(i);
    } else {
      root = i;
    }
  }
  if (n % 2 == 0) {
    cout << "NO"
         << "\n";
  } else {
    dfs(root);
    cout << "YES"
         << "\n";
    for (int x : res) {
      cout << x << " ";
    }
  }
  return 0;
}

```