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 = 2e5 + 9;
int n, tr[N], de[N], cnt, ans[N], eid, p[N];
struct edge {
  int v, nxt;
} e[N << 1];
bool vis[N];
void add(int u, int v) {
  eid++, de[v]++;
  e[eid].v = v, e[eid].nxt = p[u], p[u] = eid;
}
void out(int u, int fa) {
  ans[++cnt] = u;
  int ccnt = cnt;
  vis[u] = true;
  for (int i = p[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v != fa && !vis[v]) out(v, u);
  }
}
void dfs(int u, int fa) {
  int child = 0;
  for (int i = p[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v != fa) dfs(v, u), child += (!vis[v]);
  }
  if (de[u] % 2 == 0 && !vis[u]) out(u, fa), de[fa]--;
}
int main() {
  scanf("%d", &n);
  int now, x;
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &x);
    if (x == 0)
      now = i;
    else
      add(i, x), add(x, i);
  }
  dfs(now, 0);
  if (cnt != n)
    printf("NO");
  else {
    printf("YES\n");
    for (int i = (1); i <= (n); i++) printf("%d\n", ans[i]);
  }
  return 0;
}