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;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 11;
int N, M, T, P, Q;
list<int> G[maxn];
bool ok = 1;
int deg[maxn];
bool dele[maxn];
list<int> ans;
void del(int u, int pu) {
  if (dele[u]) return;
  dele[u] = 1;
  ans.push_back(u);
  for (int v : G[u]) {
    if (v == pu) continue;
    del(v, u);
  }
}
void dfs(int u = 1, int pu = -1) {
  for (int v : G[u]) {
    if (v == pu) continue;
    dfs(v, u);
    if (dele[v]) deg[u]--;
  }
  if (deg[u] % 2 == 0) del(u, pu);
}
int main() {
  cin >> N;
  for (int i = 1, j; i <= N; i++) {
    scanf("%d", &j);
    if (j) G[i].push_back(j), G[j].push_back(i);
  }
  for (int i = 1; i <= N; i++) deg[i] = G[i].size();
  dfs();
  if (ans.size() < N)
    puts("NO");
  else {
    puts("YES");
    for (int x : ans) printf("%d\n", x);
  }
  return 0;
}
