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 arr[1000050], d[1000050], tim[1000050], ind[1000050];
set<int> e[1000050];
vector<int> g[1000050];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    q;
int main() {
  int n, i, x, vs, curx, curd, next, size;
  pair<int, int> pa;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x) {
      e[x].insert(i);
      e[i].insert(x);
      d[x]++;
      d[i]++;
    }
  }
  for (i = 1; i <= n; i++) q.push(make_pair(d[i], i));
  for (; !q.empty();) {
    pa = q.top();
    curx = pa.second;
    curd = pa.first;
    q.pop();
    if (d[curx] != curd) continue;
    arr[++vs] = curx;
    if (d[curx]) {
      next = *e[curx].begin();
      d[curx]--;
      d[next]--;
      e[curx].erase(next);
      e[next].erase(curx);
      q.push(make_pair(d[next], next));
      if (tim[curx] % 2 == 0) {
        tim[next]++;
        g[next].push_back(curx);
        ind[curx]++;
      } else {
        g[curx].push_back(next);
        ind[next]++;
      }
    } else if (tim[curx] & 1) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
  for (; !q.empty(); q.pop())
    ;
  for (i = 1; i <= n; i++) q.push(make_pair(ind[i], i));
  for (; !q.empty();) {
    pa = q.top();
    curx = pa.second;
    curd = pa.first;
    q.pop();
    if (ind[curx] != curd) continue;
    printf("%d\n", curx);
    size = g[curx].size();
    for (i = 0; i < size; i++) {
      ind[g[curx][i]]--;
      q.push(make_pair(ind[g[curx][i]], g[curx][i]));
    }
  }
  return 0;
}