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;
template <class T>
inline bool read(T &num) {
  char in;
  bool IsN = false;
  while (in = getchar(), in != '-' && (in < '0' || in > '9'))
    if (in == EOF) return false;
  in == '-' ? (IsN = true, num = 0) : num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') num = num * 10 + in - '0';
  return IsN ? num = -num : num, true;
}
const int maxn = 2e5 + 7;
const int inf = 2e9 + 7;
const int mod = 1e9 + 7;
stack<int> tree;
vector<int> edge[maxn];
int deg[maxn];
vector<int> ans;
int fa[maxn], vis[maxn];
void dfs(int u, int p) {
  tree.push(u);
  fa[u] = p;
  for (int i = (0); i < (edge[u].size()); i++) {
    if (edge[u][i] != p) dfs(edge[u][i], u);
  }
}
void gao(int u) {
  ans.push_back(u);
  vis[u] = 1;
  for (int i = (0); i < (edge[u].size()); i++) {
    int v = edge[u][i];
    deg[v]--;
    if (v == fa[u]) continue;
    if (vis[v]) continue;
    if (deg[v] % 2 == 0) gao(v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int root;
  for (int i = (1); i < (n + 1); i++) {
    int k;
    cin >> k;
    if (k) {
      edge[i].push_back(k);
      edge[k].push_back(i);
      deg[k]++;
      deg[i]++;
    } else
      root = i;
  }
  dfs(root, -1);
  while (!tree.empty()) {
    int u = tree.top();
    tree.pop();
    if (deg[u] % 2 == 0) {
      gao(u);
    }
  }
  if (ans.size() == n) {
    cout << "YES" << endl;
    for (int i = (0); i < (ans.size()); i++) cout << ans[i] << endl;
  } else
    cout << "NO" << endl;
  return 0;
}