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;
long long n, rt, pr[200069], sq[200069], zs = 0;
vector<long long> al[200069];
bitset<200069> dg, vtd;
void trv(long long x) {
  long long i, sz = al[x].size(), l;
  vtd[x] = 1;
  zs++;
  sq[zs] = x;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    if (!vtd[l]) {
      trv(l);
    }
  }
}
void dfs(long long x) {
  long long i, sz = al[x].size(), l;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    dfs(l);
  }
  if (!dg[x]) {
    trv(x);
    dg[pr[x]] = !dg[pr[x]];
  }
}
int main() {
  long long i;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", pr + i);
    if (!pr[i]) {
      rt = i;
    } else {
      al[pr[i]].push_back(i);
      dg[i] = !dg[i];
      dg[pr[i]] = !dg[pr[i]];
    }
  }
  dfs(rt);
  if (zs < n) {
    printf("NO\n");
  } else {
    printf("YES\n");
    for (i = 1; i <= zs; i++) {
      printf("%lld\n", sq[i]);
    }
  }
}