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;
struct Edge {
  int to;
  int next;
  int val;
} edge[400005];
int head[200005];
int cnt = 1;
void addedge(int u, int v) {
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
int rt;
int num[200005];
bool vis[200005];
int ans[200005];
int tot = 0;
void dfs1(int u, int fa) {
  ans[++tot] = u;
  vis[u] = 1;
  for (int i = head[u]; i; i = edge[i].next) {
    int to = edge[i].to;
    if (to == fa) continue;
    if (!vis[to]) dfs1(to, u);
  }
}
void dfs(int u, int fa) {
  for (int i = head[u]; i; i = edge[i].next) {
    int to = edge[i].to;
    if (to == fa) continue;
    dfs(to, u);
    if (vis[to]) num[u] ^= 1;
  }
  if (!num[u]) dfs1(u, fa);
}
set<int> s0;
set<int> s1;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x) {
      addedge(i, x);
      addedge(x, i);
    } else
      rt = i;
  }
  dfs(rt, 0);
  if (tot < n) return printf("NO\n"), 0;
  printf("YES\n");
  for (int i = tot; i >= 1; i--) printf("%d\n", ans[i]);
}
