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 = 2e5 + 5;
vector<int> G[maxn];
void insert(int u, int v) {
  G[u].push_back(v);
  G[v].push_back(u);
}
int sz[maxn];
void dfs(int u, int fa) {
  sz[u] = 1;
  for (int v : G[u]) {
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
  }
}
void del(int u, int fa) {
  for (int v : G[u]) {
    if (v == fa) continue;
    if (sz[v] % 2 == 0) del(v, u);
  }
  printf("%d\n", u);
  for (int v : G[u]) {
    if (v == fa) continue;
    if (sz[v] & 1) del(v, u);
  }
}
int n;
int main() {
  scanf("%d", &n);
  int rt;
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    if (x == 0)
      rt = i;
    else
      insert(i, x);
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  dfs(rt, 0);
  del(rt, 0);
  return 0;
}