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;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
void func(void) {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
void printStack(stack<string> s) {
  if (s.empty()) return;
  string x = s.top();
  s.pop();
  printStack(s);
  cout << x;
  s.push(x);
}
long long binaryexp(long long n, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0)
    return binaryexp((n % 1000000009) * (n % 1000000009), p / 2);
  else
    return ((n % 1000000009) *
            binaryexp(((n % 1000000009) * (n % 1000000009)) % 1000000009,
                      (p - 1) / 2) %
            1000000009);
}
vector<long long> spf(10000001, 0);
void siv() {
  spf[1] = 1;
  for (long long i = 2; i < 10000001; i++) spf[i] = i;
  for (long long i = 4; i < 10000001; i += 2) {
    spf[i] = 2;
  }
  for (long long i = 3; i * i < 10000001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 10000001; j += i) {
        if (spf[j] == j) {
          spf[j] = i;
        }
      }
    }
  }
}
long long root[5007];
long long a[5007];
long long max1 = 0;
void findroot(long long i) {
  long long c1 = 0;
  while (i != root[i]) {
    c1++;
    i = root[i];
  }
  max1 = max(c1, max1);
}
int main() {
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long temp;
    cin >> temp;
    if (temp != -1) root[i] = temp;
  }
  for (int i = 1; i <= n; i++) findroot(i);
  cout << max1 << endl;
}
