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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
using namespace std;
const long long maxn = 5e5 + 123, N = 500, mod = 1e9 + 9;
vector<int> g[maxn];
int n, m, dp[maxn][2];
vector<int> ans;
void dfs(int v, int p) {
  vector<int> v1, v2, v3;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) {
      dfs(to, v);
      if (dp[to][0] && !dp[to][1]) v1.push_back(to);
      if (dp[to][1] && !dp[to][0]) v2.push_back(to);
      if (dp[to][0] && dp[to][1]) v3.push_back(to);
    }
  }
  if ((v1.size() % 2 == 0 || v3.size() > 0)) dp[v][0] = 1;
  if (v != 1 && (v1.size() % 2 == 1 || v3.size() > 0)) dp[v][1] = 1;
}
void dfs2(int v, int p, int hp) {
  vector<int> v1, v2, v3;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) {
      if (dp[to][0] && !dp[to][1]) v1.push_back(to);
      if (dp[to][1] && !dp[to][0]) v2.push_back(to);
      if (dp[to][0] && dp[to][1]) v3.push_back(to);
    }
  }
  if (v1.size() % 2 == 1 - hp) {
    v1.push_back(v3.back());
    v3.pop_back();
  }
  if (hp == 0) {
    for (int i = 0; i < v2.size(); i++) dfs2(v2[i], v, 1);
    for (int i = 0; i < v3.size(); i++) dfs2(v3[i], v, 1);
    cout << v << endl;
    for (int i = 0; i < v1.size(); i++) dfs2(v1[i], v, 0);
  } else {
    for (int i = 0; i < v2.size(); i++) dfs2(v2[i], v, 1);
    for (int i = 0; i < v3.size(); i++) dfs2(v3[i], v, 1);
    cout << v << endl;
    for (int i = 0; i < v1.size(); i++) dfs2(v1[i], v, 0);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int v;
    cin >> v;
    if (v != 0) {
      g[v].push_back(i);
      g[i].push_back(v);
    }
  }
  dfs(1, 1);
  if (dp[1][0] == 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  dfs2(1, 1, 0);
}
