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 int N = 2e5 + 5;
int n, root;
vector<int> adj[N];
int dp[N][2], deg[N];
bool mark[N];
void dfs(int u, int pre) {
  for (int v : adj[u]) {
    if (v == pre) continue;
    dfs(v, u);
  }
  for (int p = 0, _n = (2); p < _n; p++) {
    int d = (deg[u] + ((p == 0 && pre != -1) ? -1 : 0));
    vector<int> ban;
    for (int v : adj[u])
      if (v != pre) {
        if (!dp[v][0]) ban.push_back(v);
      }
    bool flag = false;
    for (int v : ban) {
      if (!dp[v][1]) {
        flag = true;
        break;
      }
    }
    if (flag) continue;
    if ((d - ban.size()) % 2 == 1) {
      flag = true;
      for (int v : adj[u])
        if (v != pre && dp[v][0]) {
          if (dp[v][1]) flag = false;
        }
      if (flag) continue;
    }
    dp[u][p] = 1;
  }
}
void print_ans(int u, int pre, int p) {
  int d = (deg[u] + ((p == 0 && pre != -1) ? -1 : 0)), cnt = 0;
  for (int v : adj[u])
    if (v != pre) {
      if (!dp[v][0]) {
        cnt++;
        print_ans(v, u, 1);
        mark[v] = 1;
      }
    }
  if ((d - cnt) % 2 == 1) {
    for (int v : adj[u])
      if (v != pre && dp[v][0] && dp[v][1]) {
        print_ans(v, u, 1);
        mark[v] = 1;
        break;
      }
  }
  cout << u << '\n';
  for (int v : adj[u])
    if (v != pre && !mark[v]) print_ans(v, u, 0);
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  clock_t tStart = clock();
  cin >> n;
  for (int i = (1), _b = (n); i <= _b; i++) {
    int j;
    cin >> j;
    if (j == 0)
      root = i;
    else {
      adj[i].push_back(j);
      adj[j].push_back(i);
      deg[i]++;
      deg[j]++;
    }
  }
  dfs(root, -1);
  if (!dp[root][0]) {
    cout << "NO" << '\n';
    return 0;
  } else {
    cout << "YES" << '\n';
    print_ans(root, -1, 0);
  }
  fprintf(stderr, "Time taken: %.2fs\n",
          (double)(clock() - tStart) / CLOCKS_PER_SEC);
  return 0;
}