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.

<image>
#include <bits/stdc++.h>
using namespace std;
vector<long long> v[200000];
vector<long long> w;
long long f[200000];
stack<long long> s;
void dfs(long long x, long long p) {
  for (auto y : v[x])
    if (y != p) dfs(y, x);
  if (f[x] % 2 == 0) {
    w.push_back(x + 1), f[x] = 0;
    for (auto y : v[x]) f[y]--;
  } else
    s.push(x + 1);
}
int main() {
  long long n, i, x;
  memset(f, 0, sizeof(f));
  scanf("%lld", &n);
  for (i = 0; i <= n - 1; i++) {
    scanf("%lld", &x);
    if (x != 0) x--, v[x].push_back(i), v[i].push_back(x), f[x]++, f[i]++;
  }
  dfs(0, -1);
  if (f[0] % 2 == 1)
    printf("NO\n");
  else {
    printf("YES\n");
    for (auto x : w) printf("%lld\n", x);
    while (!s.empty()) printf("%lld\n", s.top()), s.pop();
  }
}
