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> v[200009];
vector<int> vv;
deque<int> dq;
bool be_ve[200009], af_ve[200009];
void dfs(int i, int p) {
  int be = 0, af = 0;
  for (auto j : v[i]) {
    if (j != p) {
      dfs(j, i);
      if (be_ve[j]) {
        be++;
      } else {
        af++;
      }
    }
  }
  if (af % 2 == 0) {
    af_ve[i] = 1;
  } else {
    be_ve[i] = 1;
  }
}
void dfs_a(int i, int p) {
  int be = 0, af = 0;
  for (auto j : v[i]) {
    if (j != p) {
      if (be_ve[j]) {
        dq.push_front(j);
      } else {
        dq.push_back(j);
      }
      dfs_a(j, i);
    }
  }
}
int main() {
  int n, i, j, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x != 0) {
      v[i].push_back(x);
      v[x].push_back(i);
    }
  }
  dq.push_back(1);
  dfs(1, 0);
  if (af_ve[1]) {
    dfs_a(1, 0);
    cout << "YES"
         << " ";
    ;
    cout << endl;
    for (auto j : dq) {
      printf("%d\n", j);
      ;
    }
  } else {
    cout << "NO"
         << " ";
    ;
  }
}