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 + 1000;
int du[N];
int fa[N];
vector<int> g[N];
queue<int> q;
bool vis[N];
void del(int u) {
  if (vis[u]) return;
  q.push(u);
  vis[u] = 1;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (vis[v]) continue;
    del(v);
  }
}
void dfs(int u) {
  for (int i = 0; i < g[u].size(); ++i) {
    dfs(g[u][i]);
  }
  if (du[u] % 2 == 0) {
    du[fa[u]]--;
    del(u);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int root;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &fa[i]);
    g[fa[i]].push_back(i);
    if (fa[i] == 0) {
      root = i;
      continue;
    }
    du[fa[i]]++;
    du[i]++;
  }
  dfs(root);
  if (q.size() != n) {
    puts("NO");
    return 0;
  }
  puts("YES");
  while (!q.empty()) {
    int v = q.front();
    printf("%d\n", v);
    q.pop();
  }
  return 0;
}