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 leaf[100009];
vector<long long> adj[10009];
map<long long, long long> m;
bool visited[10009];
void initialize2(long long n) {
  for (int i = 1; i <= n; i++) {
    leaf[i] = false;
  }
}
void initialize(long long n) {
  for (int i = 1; i <= n; i++) {
    visited[i] = false;
  }
}
void dfs(long long src, long long n) {
  stack<long long> st;
  visited[src] = true;
  st.push(src);
  int flag = 1;
  int flag2 = 0;
  long long u = src;
  while (!st.empty()) {
    if (0 < m[u] && m[u] < 3 && flag == 0) {
      flag2 = 1;
      break;
    } else if (m[u] == 0 && leaf[u] == true) {
      flag2 = 1;
      break;
    } else {
      u = st.top();
    }
    st.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      flag = 0;
      if (visited[adj[u][i]] == false) {
        if (m[adj[u][i]] > 0) m[u]--;
        visited[adj[u][i]] = true;
        st.push(adj[u][i]);
      }
    }
  }
  if (flag2 == 1)
    cout << "No" << endl;
  else
    cout << "Yes" << endl;
}
long long level[10009];
long long last_element;
bool visited4[100009];
void initialize3(long long n) {
  for (int i = 1; i <= n; i++) {
    visited4[i] = false;
  }
}
void bfs(long long src, long long n) {
  initialize3(n);
  visited[src] = true;
  visited4[src] = true;
  level[src] = 1;
  last_element = src;
  queue<long long> q;
  q.push(src);
  int flag = 0;
  while (!q.empty()) {
    long long u = q.front();
    flag = 1;
    q.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      if (visited4[adj[u][i]] == false) {
        last_element = adj[u][i];
        level[adj[u][i]] = level[u] + 1;
        flag = 0;
        visited[adj[u][i]] = true;
        visited4[src] = true;
        q.push(adj[u][i]);
      }
    }
  }
}
int main() {
  long long n;
  cin >> n;
  initialize2(n);
  long long y;
  int flag = 1;
  int src;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    if (x != -1) {
      if (flag == 1) src = x;
      adj[x].push_back(i);
    }
  }
  initialize(n);
  long long last_level = INT_MIN;
  for (int i = 1; i <= n; i++) {
    if (visited[i] == false) {
      bfs(i, n);
      last_level = max(last_level, level[last_element]);
    }
  }
  cout << last_level << endl;
}
