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;
const int MAXN = 220000;
int vis[MAXN], ans[MAXN];
struct Edge {
  int u, v, nxt;
} e[MAXN << 1];
int e_cnt;
int head[MAXN];
void Add(int u, int v) {
  int id = ++e_cnt;
  e[id].u = u;
  e[id].v = v;
  e[id].nxt = head[u];
  head[u] = id;
}
void DFS2(int u, int fa) {
  ans[++ans[0]] = u;
  vis[u] = true;
  for (int id = head[u]; ~id; id = e[id].nxt) {
    int v = e[id].v;
    if (v == fa) continue;
    if (!vis[v]) DFS2(v, u);
  }
}
void DFS1(int u, int fa) {
  int siz = (u != 1);
  for (int id = head[u]; ~id; id = e[id].nxt) {
    int v = e[id].v;
    if (v == fa) continue;
    DFS1(v, u);
    if (!vis[v]) siz++;
  }
  if (siz % 2 == 0) DFS2(u, fa);
}
int main() {
  memset(head, -1, sizeof(head));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int j;
    scanf("%d", &j);
    if (j) Add(j, i), Add(i, j);
  }
  DFS1(1, 0);
  if (ans[0] == n) {
    printf("YES\n");
    for (int i = 1; i <= ans[0]; ++i) printf("%d\n", ans[i]);
  } else
    printf("NO\n");
  return 0;
}