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 double eps = 1e-9;
const int siz = 2e5 + 5, siz2 = 1e5 + 5;
int n, parent[siz], degree[siz], still[siz];
vector<int> adj[siz], ans, child[siz], ready;
bool vis[siz], del[siz];
void root(int i) {
  vis[i] = true;
  for (auto v : adj[i]) {
    if (vis[v]) continue;
    child[i].push_back(v);
    parent[v] = i;
    degree[i]++;
    degree[v]++;
    root(v);
  }
  int sz = child[i].size();
  still[i] = sz;
  if (sz == 0) {
    ready.push_back(i);
  }
}
void propagate(int i) {
  del[i] = true;
  ans.push_back(i);
  for (auto v : child[i]) {
    if (!del[v]) {
      propagate(v);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (v != 0) {
      adj[v].push_back(i);
      adj[i].push_back(v);
    }
  }
  if (n == 1) {
    printf("YES\n1\n");
    return 0;
  }
  root(1);
  while (!ready.empty()) {
    int cur = ready.back();
    ready.pop_back();
    still[parent[cur]]--;
    if (still[parent[cur]] == 0) {
      ready.push_back(parent[cur]);
    }
    if (degree[cur] % 2 == 0) {
      degree[parent[cur]]--;
      propagate(cur);
    }
  }
  int sz = ans.size();
  if (sz == n) {
    printf("YES\n");
    for (auto v : ans) {
      printf("%d\n", v);
    }
  } else {
    printf("NO\n");
  }
  return 0;
}