#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, root, deg[N], fa[N], del[N];
vector<int> g[N], ans;
void dfs2(int x) {
  ans.push_back(x);
  del[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int v = g[x][i];
    if (!del[v]) dfs2(v);
  }
}
void dfs(int x) {
  for (int i = 0; i < g[x].size(); ++i) {
    int v = g[x][i];
    dfs(v);
    if (!del[v]) deg[x] ^= 1;
  }
  if (fa[x]) deg[x] ^= 1;
  if (!deg[x]) dfs2(x);
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &fa[i]);
    if (!fa[i])
      root = i;
    else
      g[fa[i]].push_back(i);
    ;
  }
  dfs(root);
  if (ans.size() < n) {
    puts("NO");
  } else {
    puts("YES");
    for (i = 0; i < n; ++i) printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following 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.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, root, deg[N], fa[N], del[N];
vector<int> g[N], ans;
void dfs2(int x) {
  ans.push_back(x);
  del[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int v = g[x][i];
    if (!del[v]) dfs2(v);
  }
}
void dfs(int x) {
  for (int i = 0; i < g[x].size(); ++i) {
    int v = g[x][i];
    dfs(v);
    if (!del[v]) deg[x] ^= 1;
  }
  if (fa[x]) deg[x] ^= 1;
  if (!deg[x]) dfs2(x);
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &fa[i]);
    if (!fa[i])
      root = i;
    else
      g[fa[i]].push_back(i);
    ;
  }
  dfs(root);
  if (ans.size() < n) {
    puts("NO");
  } else {
    puts("YES");
    for (i = 0; i < n; ++i) printf("%d\n", ans[i]);
  }
  return 0;
}

```