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 = 1e6 + 7;
const int M = 100 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 9;
vector<int> edge[N], e[N], ans;
int n, cnt[N], deg[N];
int vis[N];
void dfs(int u, int pre) {
  cnt[u] = 1;
  for (int v : edge[u]) {
    if (v == pre) continue;
    dfs(v, u);
    cnt[u] += cnt[v];
  }
}
bool dfs2(int u) {
  vis[u] = -1;
  for (int v : e[u]) {
    if (vis[v] == -1) return false;
    if (!vis[v] && !dfs2(v)) return false;
  }
  vis[u] = 1;
  ans.push_back(u);
  return true;
}
void dfs3(int u, int pre) {
  for (int v : edge[u]) {
    if (v == pre) continue;
    if (cnt[v] % 2) {
      e[u].push_back(v);
      deg[v]++;
    } else {
      e[v].push_back(u);
      deg[u]++;
    }
    dfs3(v, u);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int fa;
    scanf("%d", &fa);
    if (fa) {
      edge[fa].push_back(i);
      edge[i].push_back(fa);
    }
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  dfs(1, 0);
  dfs3(1, 0);
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (!dfs2(i)) {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  for (int i = ans.size() - 1; i >= 0; i--) printf("%d\n", ans[i]);
  return 0;
}