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 long long INF = 1e9 + 9;
const int N = 2e5 + 10;
vector<int> v[N];
vector<int> v2[N];
int sons[N];
int flag[N];
vector<int> ans;
void dfs2(int x) {
  ans.push_back(x);
  for (auto &y : v2[x]) dfs2(y);
}
bool dfs(int x, int p = 0) {
  if (p) sons[x] = 1;
  for (auto &y : v[x]) {
    dfs(y, x);
    sons[x]++;
  }
  flag[x] = sons[x] & 1;
  if (flag[x] == 0) {
    ans.push_back(x);
    for (auto &y : v2[x]) {
      dfs2(y);
    }
    sons[p]--;
  } else {
    if (p == 0)
      return false;
    else
      v2[p].push_back(x);
  }
  return true;
}
int main() {
  int n;
  cin >> n;
  int rt = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    v[x].push_back(i);
  }
  if (dfs(v[0].back())) {
    puts("YES");
    for (auto &x : ans) printf("%d\n", x);
  } else
    puts("NO");
  return 0;
}