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;
int n, first[200005], tot, x, con[200005], pap[200005], vis[200005];
int ans[200005];
struct data {
  int u, v, next;
} s[400005];
void add(int x, int y) {
  s[++tot].u = x;
  s[tot].v = y;
  s[tot].next = first[x];
  first[x] = tot;
}
stack<int> q;
void build(int x, int pre) {
  q.push(x);
  pap[x] = pre;
  for (int i = first[x]; i; i = s[i].next) {
    int j = s[i].v;
    if (j == pre) continue;
    build(j, x);
  }
}
void dfs(int x) {
  ans[++ans[0]] = x;
  vis[x] = 1;
  for (int i = first[x]; i; i = s[i].next) {
    int j = s[i].v;
    con[j]--;
    if (pap[x] == j) continue;
    if (vis[j]) continue;
    if (con[j] % 2 == 0) dfs(j);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (!x) continue;
    add(i, x);
    add(x, i);
    con[x]++;
    con[i]++;
  }
  build(1, 0);
  while (q.size()) {
    int xx = q.top();
    q.pop();
    if (con[xx] % 2 == 0) dfs(xx);
  }
  if (ans[0] == n) {
    printf("YES\n");
    for (int i = 1; i <= ans[0]; ++i) printf("%d\n", ans[i]);
  } else
    printf("NO");
}
