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;
int main() {
  int node_num;
  while (scanf("%d", &node_num) == 1) {
    vector<int> G[2001];
    int a[2001];
    memset(a, 0, sizeof(a));
    int distance[2001];
    for (int i = int(1); i <= int(node_num); i++) {
      int n1 = i;
      int n2;
      scanf("%d", &n2);
      if (n2 != -1)
        G[n2].push_back(n1);
      else {
        a[n1] = -1;
        distance[n1] = 1;
      }
    }
    int maxx = 1;
    int taken[2001];
    memset(taken, 0, sizeof(taken));
    for (int i = int(1); i <= int(node_num); i++) {
      if ((a[i] == -1) && !taken[i]) {
        queue<int> q;
        q.push(i);
        taken[i] = 1;
        while (!q.empty()) {
          int u = q.front();
          q.pop();
          for (int j = int(0); j <= int(G[u].size() - 1); j++) {
            int v = G[u][j];
            if (!taken[v]) {
              taken[v] = 1;
              distance[v] = distance[u] + 1;
              maxx = max(maxx, distance[v]);
              q.push(v);
            }
          }
        }
      }
    }
    printf("%d\n", maxx);
  }
  return 0;
}
