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 = 200000 + 9;
int n, p[N], deg[N];
vector<int> G[N];
unordered_set<int> ust[N][2];
int dfs(int u, int p) {
  int sz = 1;
  for (int v : G[u])
    if (v != p) {
      int tmp = dfs(v, u);
      ust[u][tmp & 1].insert(v);
      sz += tmp;
    }
  if (p) ust[u][(n - sz) & 1].insert(p);
  return sz;
}
bool vis[N];
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    if (!p[i]) continue;
    G[p[i]].push_back(i);
    G[i].push_back(p[i]);
  }
  dfs(1, 0);
  queue<int> Q;
  vector<int> vc;
  for (int i = 1; i <= n; ++i) {
    if (ust[i][0].empty() && (~ust[i][1].size() & 1u)) {
      Q.push(i);
      vis[i] = true;
    }
  }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    vc.push_back(u);
    for (int v : G[u])
      if (!vis[v]) {
        ust[v][0].erase(u);
        ust[v][1].erase(u);
        if (ust[v][0].empty() && (~ust[v][1].size() & 1u)) {
          Q.push(v);
          vis[v] = true;
        }
      }
  }
  if ((int)vc.size() == n) {
    printf("YES\n");
    for (int u : vc) printf("%d\n", u);
  } else {
    printf("NO\n");
  }
}
int main() {
  int T = 1;
  while (T--) solve();
  return 0;
}