#include <bits/stdc++.h>
using namespace std;
const int maxn = 2 * (int)(1e5) + 10;
const int inf = 0x3f3f3f3f;
const int mod = (int)1e9 + 9;
const double eps = 1e-6;
vector<int> edge[maxn * 2];
int sz[maxn];
void dfs1(int u) {
  sz[u] = 1;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    dfs1(v);
    sz[u] += sz[v];
  }
}
void dfs2(int u) {
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (sz[v] % 2 == 0) dfs2(v);
  }
  printf("%d\n", u);
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (sz[v] % 2) dfs2(v);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) edge[i].clear();
    int root, v;
    for (int u = 1; u <= n; u++) {
      scanf("%d", &v);
      if (!v) root = u;
      edge[v].push_back(u);
    }
    if (n % 2 == 0) {
      printf("NO\n");
      continue;
    } else {
      memset(sz, 0, sizeof(sz));
      printf("YES\n");
      dfs1(root);
      dfs2(root);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 2 * (int)(1e5) + 10;
const int inf = 0x3f3f3f3f;
const int mod = (int)1e9 + 9;
const double eps = 1e-6;
vector<int> edge[maxn * 2];
int sz[maxn];
void dfs1(int u) {
  sz[u] = 1;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    dfs1(v);
    sz[u] += sz[v];
  }
}
void dfs2(int u) {
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (sz[v] % 2 == 0) dfs2(v);
  }
  printf("%d\n", u);
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (sz[v] % 2) dfs2(v);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) edge[i].clear();
    int root, v;
    for (int u = 1; u <= n; u++) {
      scanf("%d", &v);
      if (!v) root = u;
      edge[v].push_back(u);
    }
    if (n % 2 == 0) {
      printf("NO\n");
      continue;
    } else {
      memset(sz, 0, sizeof(sz));
      printf("YES\n");
      dfs1(root);
      dfs2(root);
    }
  }
  return 0;
}

```