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 = 4e5 + 10;
const int inf = 1e9;
int h[N], e[N], ne[N], idx;
int st[N];
int vis[N], f[N];
vector<int> ans;
int sign;
void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
void dfs(int u, int fa) {
  int i;
  int cnt = 0;
  for (i = h[u]; i != -1; i = ne[i]) {
    int j = e[i];
    if (j == fa) continue;
    f[j] = u;
    dfs(j, u);
    if (vis[j]) {
      continue;
    } else {
      cnt++;
    }
  }
  if (cnt % 2 && u != 1) {
    vis[u] = 1;
    ans.push_back(u);
  }
  if (u == 1) {
    sign = cnt;
  }
}
void dfs1(int u, int fa) {
  int i;
  for (i = h[u]; i != -1; i = ne[i]) {
    int j = e[i];
    if (j == fa) continue;
    if (vis[f[j]] && !vis[j]) {
      ans.push_back(j);
      vis[j] = 1;
    }
    dfs1(j, u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  memset(h, -1, sizeof h);
  int i;
  int n;
  cin >> n;
  for (i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x != 0) {
      add(x, i);
      add(i, x);
    }
  }
  dfs(1, 0);
  if (!vis[1] && sign % 2 == 0) {
    ans.push_back(1);
    vis[1] = 1;
  }
  dfs1(1, 0);
  if ((int)ans.size() < n) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (auto x : ans) {
      cout << x << endl;
    }
  }
  return 0;
}