Party

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

Copy

5
-1
1
2
1
-1

output

Copy

3

Note

For the first example, three groups are sufficient, for example:

• Employee 1
• Employees 2 and 4
• Employees 3 and 5

---

刚开始是没明白题，直接把每个节点拿来暴搜，妥妥TLE。把-1作为头开始搜更有效

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#define ll long long
using namespace std;
int n,depth=0;
vector<int> p,head;
map<int, vector<int>> mp;
void dfs(int i,int depth,int &res)
{
    depth++;
    res = max(depth, res);
    if (mp[i].size() == 0)
    {
        return;
    }
    for (int k = 0; k < mp[i].size(); k++)
    {
        dfs(mp[i][k], depth, res);
    }
}
int main()
{
    cin >> n;
    p.resize(n+1);
    
    for (int i = 1; i < n+1; i++)
    {
        cin >> p[i];
        if (mp.find(i) == mp.end())
        {
            mp[i] = vector<int>();
        }
        if (p[i] != -1)
        {
            if (mp.find(p[i]) == mp.end())
            {
                mp[p[i]] = vector<int>();
                mp[p[i]].push_back(i);
            }
            else
            {
                mp[p[i]].push_back(i);
            }
        }
        else
        {
            head.push_back(i);
        }
    }
    int res = 0;
    for (int i = 0; i <head.size() ; i++)
    {
            dfs(head[i], 0, res);
    }
    cout << res;
    //system("pause");
    return 0;
}

贴给别人写的，不用邻接表。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
int n,a[2010],ans,tem;
void dfs(int u)
{
    tem++;
    if(a[u]==-1)
        return ;
    dfs(a[u]);
}
void solve()
{
    ans=-INF;
    for(int i=1;i<=n;i++)
    {
        tem=0;
        dfs(i);
        ans=max(ans,tem);
    }
    printf("%d
",ans);
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        solve();
    }
    return 0;
}