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> > tree;
vector<int> stopnie;
vector<int> dist;
vector<int> ans;
vector<bool> deleted;
bool bad = false;
void clearDown(int cur, int p) {
  if (stopnie[cur] % 2 == 1) {
    bad = true;
  }
  ans.push_back(cur);
  deleted[cur] = true;
  for (auto a : tree[cur]) {
    if (a != p && deleted[a] == false) {
      stopnie[a]--;
      clearDown(a, cur);
    }
  }
}
void dfs(int cur, int p) {
  for (auto a : tree[cur]) {
    if (a == p) continue;
    dfs(a, cur);
  }
  if (stopnie[cur] % 2 == 0) {
    ans.push_back(cur);
    deleted[cur] = true;
    stopnie[cur] = 0;
    for (auto a : tree[cur]) {
      if (deleted[a] == true) continue;
      stopnie[a]--;
      if (a != p) clearDown(a, cur);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  tree.resize(n);
  stopnie.resize(n);
  dist.resize(n);
  deleted.resize(n);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a > 0) {
      tree[i].push_back(a - 1);
      tree[a - 1].push_back(i);
      stopnie[i]++;
      stopnie[a - 1]++;
    }
  }
  if (n == 1) {
    cout << "YES\n1\n";
    return 0;
  }
  if (n == 2) {
    cout << "NO\n";
    return 0;
  }
  int root = 0;
  for (int i = 0; i < n; i++)
    if (stopnie[i] > 1) root = i;
  dfs(root, -1);
  for (auto a : deleted)
    if (a == false) bad = true;
  if (bad == true) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (auto a : ans) cout << a + 1 << "\n";
    cout << endl;
  }
}