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;
vector<vector<int>> vt;
vector<int> dap;
int degree[200010];
int visited[200010];
int n;
void solve(int here) {
  visited[here] = 1;
  dap.push_back(here);
  for (int i = 0; i < vt[here].size(); i++) {
    int next = vt[here][i];
    if (!visited[next]) solve(next);
  }
}
void dfs(int here, int par) {
  for (int i = 0; i < vt[here].size(); i++) {
    dfs(vt[here][i], here);
  }
  if (degree[here] % 2 == 0) {
    degree[par]--;
    solve(here);
  }
}
int main() {
  scanf("%d", &n);
  vt.resize(n + 1);
  int ch = 0;
  for (int i = 1; i <= n; i++) {
    int q;
    scanf("%d", &q);
    if (q == 0) {
      ch = i;
      continue;
    }
    vt[q].push_back(i);
    degree[q]++;
    degree[i]++;
  }
  if (n % 2 == 0) {
    puts("NO");
  } else {
    puts("YES");
    dfs(ch, 0);
    for (int i = 0; i < dap.size(); i++) printf("%d\n", dap[i]);
  }
}