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;
vector<int> E[200005];
int deg[200005];
bool dele[200005];
int n;
void del(int u) {
  deg[u] = 0;
  dele[u] = 1;
  printf("%d\n", u);
  for (int i = 0; i < E[u].size(); i++) {
    if (deg[E[u][i]] != 0) deg[E[u][i]]--;
  }
  return;
}
void destroy(int fa, int u) {
  del(u);
  for (int i = 0; i < E[u].size(); i++) {
    if (E[u][i] != fa && dele[E[u][i]] == 0) {
      destroy(u, E[u][i]);
    }
  }
  return;
}
void dfs(int fa, int u) {
  for (int i = 0; i < E[u].size(); i++) {
    if (E[u][i] != fa) {
      dfs(u, E[u][i]);
    }
  }
  if (deg[u] % 2 == 0 && dele[u] == 0) {
    destroy(fa, u);
  }
  return;
}
int main() {
  scanf("%d", &n);
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    int g;
    scanf("%d", &g);
    if (g) {
      E[g].push_back(i);
      E[i].push_back(g);
      deg[g]++;
      deg[i]++;
    }
  }
  dfs(0, 1);
  return 0;
}