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 SIZE = 200001;
struct tree {
  int father;
  vector<int> sons;
  bool deleted;
  int siz;
};
tree tr[SIZE];
void write(int x) {
  int y;
  for (int i = 0; i < tr[x].sons.size(); i++) {
    y = tr[x].sons[i];
    if (!tr[y].deleted && y != tr[x].father) {
      printf("%d\n", y);
      tr[y].deleted = true;
      write(y);
    }
  }
}
void alg() {
  int x, y;
  queue<int> q;
  stack<int> Q;
  q.push(1);
  Q.push(1);
  while (!q.empty()) {
    x = q.front();
    for (int i = 0; i < tr[x].sons.size(); i++) {
      y = tr[x].sons[i];
      if (tr[x].father != y) {
        tr[y].father = x;
        q.push(y);
        Q.push(y);
      }
    }
    q.pop();
  }
  while (!Q.empty()) {
    x = Q.top();
    if ((tr[x].siz & 1) == 0) {
      printf("%d\n", x);
      tr[x].deleted = true;
      tr[tr[x].father].siz--;
      write(x);
    }
    Q.pop();
  }
}
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (a != 0) {
      tr[i].sons.push_back(a);
      tr[a].sons.push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) tr[i].siz = tr[i].sons.size();
  if ((n & 1) == 0)
    printf("NO");
  else {
    printf("YES\n");
    alg();
  }
  return 0;
}