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 + 10;
vector<int> g[N];
int du[N], fa[N];
bool vis[N];
queue<int> q;
void dfs(int r, int f) {
  fa[r] = f;
  for (int i = 0; i < g[r].size(); i++) {
    int v = g[r][i];
    if (v != f) dfs(v, r);
  }
  q.push(r);
}
vector<int> ans;
void del(int x) {
  vis[x] = true;
  ans.push_back(x);
  for (int i = 0; i < g[x].size(); i++) {
    int v = g[x][i];
    if (!vis[v] && fa[x] != v) del(v);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int x;
  memset(du, 0, sizeof(du));
  memset(vis, false, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x == 0) continue;
    du[x] += 1;
    du[i] += 1;
    g[i].push_back(x);
    g[x].push_back(i);
  }
  dfs(1, -1);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (du[x] % 2 == 0) {
      del(x);
      du[fa[x]] -= 1;
    }
  }
  if (ans.size() != n)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 0; i < n; i++) printf("%d\n", ans[i]);
  }
  return 0;
}