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>
using namespace std;
struct edge {
  int v, n;
} e[400005];
int head[200005], cnt, vis[200005], deg[200005];
inline void add(int u, int v) {
  e[++cnt].v = v;
  e[cnt].n = head[u];
  head[u] = cnt;
}
inline void dfs2(int u, int fa) {
  if (vis[u]) return;
  if (u != 0) printf("%d\n", u);
  vis[u] = 1;
  for (int i = head[u]; i; i = e[i].n) {
    if (vis[e[i].v]) continue;
    if (e[i].v == fa) continue;
    dfs2(e[i].v, u);
  }
}
inline void dfs(int u, int fa) {
  if (vis[u]) return;
  for (int i = head[u]; i; i = e[i].n) {
    if (e[i].v == fa) continue;
    dfs(e[i].v, u);
  }
  if (deg[u] & 1) return;
  dfs2(u, fa);
  deg[fa]--;
}
int main(int argc, char** argv) {
  int n;
  cin >> n;
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    add(i, v);
    add(v, i);
    if (v == 0) --deg[i];
    ++deg[v], ++deg[i];
  }
  dfs(0, 0);
  return 0;
}