A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:

  * Employee A is the immediate manager of employee B
  * Employee B has an immediate manager employee C such that employee A is the superior of employee C. 



The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.

Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.

What is the minimum number of groups that must be formed?

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.

The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager. 

It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.

Output

Print a single integer denoting the minimum number of groups that will be formed in the party.

Examples

Input

5
-1
1
2
1
-1


Output

3

Note

For the first example, three groups are sufficient, for example: 

  * Employee 1 
  * Employees 2 and 4 
  * Employees 3 and 5 
#include <bits/stdc++.h>
using namespace std;
const int inf = (1 << 30);
const int N = 2005;
const int mod = 1007;
const double eps = 1e-10;
int n, m;
vector<int> e[N];
bool in[N];
int dfs(int u, int f, int dep) {
  int mx = dep;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == f) continue;
    mx = max(mx, dfs(v, u, dep + 1));
  }
  return mx;
}
int main() {
  int i, j, k, t, cas = 0;
  int a, b;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i <= n; i++) e[i].clear();
    memset(in, 0, sizeof(in));
    for (i = 1; i <= n; i++) {
      scanf("%d", &a);
      if (a == -1) continue;
      e[a].push_back(i);
      in[i] = 1;
    }
    int ret = 0;
    for (i = 1; i <= n; i++)
      if (!in[i]) ret = max(ret, dfs(i, i, 1));
    printf("%d\n", ret);
  }
  return 0;
}
