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 + 1;
vector<int> g[N], ans;
int n, done[N];
void fix(int u, int p) {
  ans.push_back(u + 1);
  done[u] = 1;
  for (auto i : g[u])
    if (i != p & !done[i]) fix(i, u);
}
bool dfs(int u, int p) {
  int even = p != -1;
  for (auto i : g[u])
    if (p != i) even ^= dfs(i, u);
  if (!even) fix(u, p);
  return even;
}
int main() {
  scanf("%d", &n);
  if (!(n & 1)) return puts("NO"), 0;
  for (int i = 0, a; i < n; i++) {
    scanf("%d", &a);
    if (a--) g[a].push_back(i), g[i].push_back(a);
  }
  dfs(0, -1);
  puts("YES");
  for (auto i : ans) printf("%d\n", i);
}