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 n, m, i, j, deg[500005], x, y, vis[500005];
vector<int> bi[500005], seq;
void dfs2(int x, int fa) {
  int i;
  vis[x] = 1;
  seq.push_back(x);
  for (i = 0; i < bi[x].size(); i++) {
    if (bi[x][i] == fa) continue;
    if (vis[bi[x][i]]) continue;
    dfs2(bi[x][i], x);
  }
}
void dfs(int x, int fa) {
  int i;
  for (i = 0; i < bi[x].size(); i++) {
    if (bi[x][i] == fa) continue;
    if (vis[bi[x][i]]) continue;
    dfs(bi[x][i], x);
  }
  if (deg[x] % 2 == 0) {
    deg[fa]--;
    vis[x] = 1;
    seq.push_back(x);
    for (i = 0; i < bi[x].size(); i++) {
      if (bi[x][i] == fa) continue;
      if (vis[bi[x][i]]) continue;
      dfs2(bi[x][i], x);
    }
  }
}
int main() {
  scanf("%d", &n);
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x == 0) continue;
    bi[x].push_back(i);
    bi[i].push_back(x);
    deg[x]++;
    deg[i]++;
  }
  dfs(1, 0);
  if (seq.size() < n) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (i = 0; i < seq.size(); i++) printf("%d\n", seq[i]);
  return 0;
}