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;
template <class Z>
inline bool checkmax(Z &a, Z b) {
  if (a == -1 || a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class Z>
inline bool checkmin(Z &a, Z b) {
  if (a == -1 || a > b) {
    a = b;
    return true;
  }
  return false;
}
int net[5100], size;
void init() {
  size = 0;
  memset(net, -1, sizeof(net));
}
struct EDGE {
  int v, next;
} edge[1100000];
void add(int u, int v) {
  edge[size].v = v;
  edge[size].next = net[u];
  net[u] = size++;
}
int deg[5100];
int DFS(int u) {
  int len = 0;
  for (int i = net[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    len = max(len, DFS(v));
  }
  return len + 1;
}
int main() {
  int i, j, h;
  int T, n, m;
  int a, b, c;
  while (scanf("%d", &n) != EOF) {
    init();
    memset(deg, 0, sizeof(deg));
    for (i = 1; i <= n; i++) {
      scanf("%d", &a);
      if (a == -1) continue;
      deg[i]++;
      add(a, i);
    }
    int ans = 0;
    for (i = 1; i <= n; i++) {
      if (deg[i] == 0) ans = max(ans, DFS(i));
    }
    printf("%d\n", ans);
  }
  return 0;
}
