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.

<image>
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
int n, tim, tmp, cnt[MAXN];
int f[MAXN], vis[MAXN], rem[MAXN];
struct node {
  int h, num;
  friend bool operator<(node xx, node yy) { return xx.h < yy.h; }
} a[MAXN];
priority_queue<node> q;
queue<int> prin;
vector<int> v[MAXN];
void dfs(int x, int h) {
  rem[x] = 1;
  int k = v[x].size();
  for (int i = 0; i < k; i++) {
    if (rem[v[x][i]]) continue;
    dfs(v[x][i], h + 1);
  }
  a[x].h = h;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tmp);
    v[tmp].push_back(i);
    if (tmp) v[i].push_back(tmp), cnt[i]++;
    cnt[tmp]++;
    a[i].num = i;
  }
  if (!(n & 1)) {
    printf("NO");
    return 0;
  }
  dfs(1, 1);
  for (int i = 1; i <= n; i++) {
    if (cnt[i] & 1) continue;
    q.push(a[i]);
  }
  while (!q.empty()) {
    node tem = q.top();
    q.pop();
    if ((cnt[tem.num] & 1) || vis[tem.num]) continue;
    vis[tem.num] = true;
    prin.push(tem.num);
    int k = v[tem.num].size();
    for (int i = 0; i < k; i++) {
      if (vis[v[tem.num][i]]) continue;
      cnt[v[tem.num][i]]--;
      if (cnt[v[tem.num][i]] & 1) continue;
      q.push(a[v[tem.num][i]]);
    }
  }
  printf("YES\n");
  while (!prin.empty()) {
    printf("%d\n", prin.front());
    prin.pop();
  }
}
