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>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
long long powermod(long long base, long long exp, long long modulus) {
  base %= modulus;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((result % modulus) * (base % modulus)) % modulus;
    base = ((base % modulus) * (base % modulus) % modulus);
    exp >>= 1;
  }
  return result;
}
using namespace std;
vector<long long> v[3000];
long long level[3000];
int main() {
  long long i, j, k, l, m, n;
  cin >> n;
  queue<long long> q;
  for (i = 1; i <= n; i++) {
    cin >> k;
    if (k == -1) {
      q.push(i);
      level[i] = 1;
    } else
      v[k].push_back(i);
  }
  long long ans = 1;
  while (!q.empty()) {
    long long u = q.front();
    q.pop();
    for (auto it : v[u]) {
      level[it] = level[u] + 1;
      ans = max(ans, level[it]);
      q.push(it);
    }
  }
  cout << ans;
}
