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 + 100;
struct compass {
  int dat, nxt;
} p[maxn * 2];
int tot, n, x, du[maxn], head[maxn], root, ha[maxn];
bool valid[maxn];
int read() {
  int tot = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
void getstr(char *a) {
  char ch = getchar();
  int len = 0;
  while (ch < 'a' || ch > 'z') ch = getchar();
  while (ch >= 'a' && ch <= 'z') {
    a[++len] = ch;
    ch = getchar();
  }
}
void connect(int x, int y) {
  tot++;
  p[tot].nxt = head[x];
  p[tot].dat = y;
  head[x] = tot;
}
void dfs(int x, int fa) {
  ha[x] = 1;
  if (valid[x]) {
    for (int i = head[x]; i; i = p[i].nxt)
      if (p[i].dat != fa && valid[p[i].dat] && ha[p[i].dat] == 0)
        dfs(p[i].dat, x);
    if (du[x] % 2 == 0) {
      cout << x << "\n";
      valid[x] = false;
      du[x] = 0;
      for (int i = head[x]; i; i = p[i].nxt)
        if (valid[p[i].dat]) du[p[i].dat]--;
      for (int i = head[x]; i; i = p[i].nxt)
        if (p[i].dat != fa && valid[p[i].dat]) dfs(p[i].dat, x);
    }
  }
}
int main() {
  n = read();
  if (n % 2 == 0) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) valid[i] = true;
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    x = read();
    if (x != 0) {
      connect(x, i);
      connect(i, x);
      du[x]++;
      du[i]++;
    } else
      root = i;
  }
  dfs(root, 0);
  for (int i = 1; i <= n; i++)
    if (valid[i]) cout << i << "\n";
  return 0;
}
