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.

<image>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 10;
vector<int> G[maxn];
bool visited[maxn];
bool dfs(int u, int fa) {
  bool bet = false;
  for (unsigned int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    visited[v] = dfs(v, u);
    if (!visited[v]) bet = !bet;
  }
  return bet;
}
void print(int u, int fa) {
  for (unsigned int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    if (visited[v]) print(v, u);
  }
  cout << u << endl;
  for (unsigned int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    if (!visited[v]) print(v, u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, rt, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x == 0)
      rt = i;
    else {
      G[x].push_back(i);
      G[i].push_back(x);
    }
  }
  memset(visited, false, sizeof(visited));
  if (!dfs(rt, 0)) {
    cout << "YES" << endl;
    print(rt, 0);
  } else
    cout << "NO" << endl;
  return 0;
}
