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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int maxn = 1e6 + 7;
int n, size[maxn], vis[maxn];
std::vector<int> ch[maxn];
void dfs1(int u, int f) {
  size[u] = 1;
  for (int v : ch[u])
    if (v != f) {
      dfs1(v, u);
      size[u] += size[v];
    }
}
void dfs(int u, int f, int dx) {
  int d = dx;
  for (int v : ch[u])
    if (v != f) d++;
  for (int v : ch[u]) {
    if (v != f && size[v] % 2 == 0) {
      dfs(v, u, 1);
      d--;
    }
  }
  if (d % 2 == 0) {
    printf("%d\n", u);
    vis[u] = 1;
    for (int v : ch[u])
      if (v != f && !vis[v]) dfs(v, u, 0);
  } else {
    int t = 0;
    for (int v : ch[u]) {
      if (v == f || vis[v]) continue;
      if (size[v] % 2 == 0) t = v;
    }
    assert(t);
    dfs(t, u, 1);
    printf("%d\n", u);
    vis[u] = 1;
    for (int v : ch[u])
      if (v != f && v != t && (!vis[v])) dfs(v, u, 0);
  }
}
int main() {
  scanf("%d", &n);
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    if (!x) continue;
    ch[i].push_back(x);
    ch[x].push_back(i);
  }
  puts("YES");
  dfs1(1, 0);
  dfs(1, 0, 0);
  return 0;
}