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>
typedef struct person {
  int num;
  int nexts[2000];
} person;
person perArr[2000];
int dfs(int count, int current) {
  int out = 0;
  int cur = 0;
  int i = 0;
  if (perArr[current].num == 0) {
    return count;
  } else {
    for (i = 0; i < perArr[current].num; i++) {
      cur = dfs(count + 1, perArr[current].nexts[i]);
      if (cur > out) {
        out = cur;
      }
    }
    return out;
  }
}
int main(int argc, char *argv[]) {
  int lines;
  scanf("%d", &lines);
  int noManager[2000] = {0};
  int count = 0;
  int i = 0;
  int c;
  for (i = 0; i < lines; i++) {
    scanf("%d", &c);
    c = c - 1;
    if (c == -2) {
      noManager[count] = i;
      count++;
    } else {
      perArr[c].nexts[perArr[c].num] = i;
      perArr[c].num++;
    }
  }
  int done = 0;
  int out = 1;
  int cur = 0;
  for (i = 0; i < count; i++) {
    cur = dfs(1, noManager[i]);
    if (cur > out) {
      out = cur;
    }
  }
  printf("%d\n", out);
  return EXIT_SUCCESS;
}
