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 N = 2e5 + 3;
vector<int> ed[N];
int dp[N][2];
int PreDel[N];
int dps(int u, int ct) {
  if (dp[u][ct] != -1) return dp[u][ct];
  int t = 0, tr = 0, sz = ed[u].size();
  int l, r;
  for (int i = 0; i < sz; i++) {
    l = dps(ed[u][i], 1);
    r = dps(ed[u][i], 0);
    if (!l && !r) return dp[u][ct] = 0;
    if (l && r) {
      PreDel[ed[u][i]] = 1;
      t = ed[u][i];
    } else if (!l && r) {
      PreDel[ed[u][i]] = 0;
      tr++;
    } else
      PreDel[ed[u][i]] = 1;
  }
  if ((tr % 2) != ct && !t) return dp[u][ct] = 0;
  if ((tr % 2) != ct) PreDel[t] = 0;
  return dp[u][ct] = 1;
}
void print(int u) {
  int sz = ed[u].size();
  for (int i = 0; i < sz; i++)
    if (PreDel[ed[u][i]] == 1) print(ed[u][i]);
  printf("%d\n", u);
  for (int i = 0; i < sz; i++)
    if (PreDel[ed[u][i]] == 0) print(ed[u][i]);
}
int main() {
  int n;
  scanf("%d", &n);
  int fr, root;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &fr);
    if (fr)
      ed[fr].push_back(i);
    else
      root = i;
  }
  memset(dp, 0xff, sizeof dp);
  int ret = dps(root, 0);
  if (!ret)
    printf("NO");
  else {
    printf("YES\n");
    print(root);
  }
  return 0;
}