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>
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
vector<int> G[N];
int vis[N], rt = 0;
void print(int x) {
  vis[x] = 1;
  printf("%d\n", x);
  for (int i = 0; i < G[x].size(); i++) {
    int v = G[x][i];
    if (!vis[v]) print(v);
  }
}
void DFS(int u, int fa) {
  int d = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (!vis[v]) DFS(v, u);
  }
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (!vis[v]) d++;
  }
  if (fa) d++;
  if (d % 2 == 0) {
    printf("%d\n", u);
    vis[u] = 1;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (!vis[v]) print(v);
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int fa;
    scanf("%d", &fa);
    if (!fa)
      rt = i;
    else
      G[fa].push_back(i);
  }
  if ((n & 1) == 0)
    printf("NO\n");
  else {
    printf("YES\n");
    DFS(rt, 0);
  }
  return 0;
}
