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;
int cnt[200000];
vector<int> e[200000];
int chk[200000];
int chk2[200000];
vector<int> ans;
queue<int> q;
void dfs(int cur) {
  chk[cur] = 1;
  int i = 0;
  while (i < e[cur].size()) {
    if (chk[e[cur][i]] == 0) dfs(e[cur][i]);
    i++;
  }
  q.push(cur);
}
int main() {
  int nsize = 0;
  scanf("%d", &nsize);
  int i = 0;
  while (i < nsize) {
    int tto;
    scanf("%d", &tto);
    if (tto != 0) {
      tto--;
      e[i].push_back(tto);
      e[tto].push_back(i);
      cnt[i]++;
      cnt[tto]++;
    }
    i++;
  }
  dfs(0);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    if ((cnt[cur] % 2) != 0 || chk2[cur] == 1) {
      continue;
    }
    chk2[cur] = 1;
    ans.push_back(cur);
    i = 0;
    while (i < e[cur].size()) {
      int next = e[cur][i];
      if (chk2[next] == 0) {
        cnt[next]--;
        q.push(next);
      }
      i++;
    }
  }
  i = 0;
  while (i < nsize) {
    if (chk2[i] == 0) break;
    i++;
  }
  if (i != nsize) {
    printf("NO\n");
  } else {
    printf("YES\n");
    i = 0;
    while (i < nsize) {
      printf("%d\n", ans[i] + 1);
      i++;
    }
  }
  return 0;
}