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 double pi = acos(-1.0);
template <class T>
T POW(T x, T y) {
  T ret = 1;
  while (y--) {
    ret *= x;
  }
  return ret;
}
template <class T>
inline bool EQUAL(T x, T y) {
  if (fabs(x - y) < 1e-9) return true;
  return false;
}
int N;
vector<int> G[2005];
bool vis[2005];
int dfs(int u) {
  vis[u] = true;
  int ret = 1;
  for (int i = 0; i < ((int)G[u].size()); i++) {
    int v = G[u][i];
    if (vis[v] == false) {
      ret += dfs(v);
    }
  }
  return ret;
}
int main() {
  int i, j, x;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%d", &x);
    if (x != -1) G[i].push_back(x);
  }
  int mx = 1, depth;
  for (i = 1; i <= N; i++) {
    if (((int)G[i].size())) {
      memset(vis, false, sizeof(vis));
      depth = dfs(i);
      mx = max(mx, depth);
    }
  }
  printf("%d\n", mx);
  return 0;
}
