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;
inline int read() {
  int x = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(int x) {
  write(x);
  puts("");
}
const int N = 2e5 + 5;
struct edge {
  int link, next;
} e[N << 1];
int n, head[N], tot, du[N];
inline void add_edge(int u, int v) {
  e[++tot] = (edge){v, head[u]};
  head[u] = tot;
}
inline void insert(int u, int v) {
  add_edge(u, v);
  add_edge(v, u);
}
inline void init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int u = read();
    if (u) insert(u, i), du[i]++, du[u]++;
  }
}
bool mark[N];
int ans[N], cnt;
void dfs(int u, int fa, bool flag) {
  if (!flag) {
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].link;
      if (v != fa && !mark[v]) {
        dfs(v, u, 0);
      }
    }
  }
  if (du[u] % 2 == 0) {
    du[u] = 0;
    ans[++cnt] = u;
    mark[u] = 1;
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].link;
      --du[v];
      if (v != fa && !mark[v]) {
        dfs(v, u, 1);
      }
    }
  }
}
inline void solve() {
  dfs(1, 0, 0);
  if (cnt == n) {
    puts("YES");
    for (int i = 1; i <= n; i++) writeln(ans[i]);
  } else {
    puts("NO");
  }
}
int main() {
  init();
  solve();
  return 0;
}
