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 mod = 1e9 + 9;
const long long inf = 0x3f3f3f3f3f3f3f;
const int maxn = 2e5 + 5;
int n, a, k, rt;
int siz[maxn];
bool vis[maxn];
vector<int> g[maxn];
vector<int> son[maxn];
void dfs1(int v) {
  siz[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    dfs1(g[v][i]);
    siz[v] += siz[g[v][i]];
    if (siz[g[v][i]] % 2 == 0) {
      son[v].push_back(g[v][i]);
    }
  }
}
void dfs(int v) {
  vis[v] = true;
  for (int i = 0; i < son[v].size(); i++) {
    dfs(son[v][i]);
  }
  printf("%d\n", v);
  for (int i = 0; i < g[v].size(); i++) {
    if (!vis[g[v][i]]) dfs(g[v][i]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &k);
    if (k == 0) rt = i;
    g[k].push_back(i);
  }
  if (n % 2 == 0) {
    printf("NO\n");
  } else {
    printf("YES\n");
    dfs1(rt);
    dfs(rt);
  }
  return 0;
}