Codeforces Round 831 (Div. 1 + Div. 2) E

Pak Chanek has n blank heart-shaped cards. Card 11 is attached directly to the wall while each of the other cards is hanging onto exactly one other card by a piece of string. Specifically, card i (i>1) is hanging onto card pi (pi<i).

In the very beginning, Pak Chanek must write one integer number on each card. He does this by choosing any permutation a of [1,2,…,n]. Then, the number written on card i is ai.

After that, Pak Chanek must do the following operation n times while maintaining a sequence s (which is initially empty):

1. Choose a card x such that no other cards are hanging onto it.
2. Append the number written on card x to the end of s.
3. If x≠1 and the number on card px is larger than the number on card x, replace the number on card px with the number on card x.
4. Remove card x.

After that, Pak Chanek will have a sequence s with n elements. What is the maximum length of the longest non-decreasing subsequence†† of s at the end if Pak Chanek does all the steps optimally?

†† A sequence b is a subsequence of a sequence c if b can be obtained from c by deletion of several (possibly, zero or all) elements. For example, [3,1] is a subsequence of [3,2,1], [4,3,1] and [3,1], but not [1,3,3,7] and [3,10,4].

Input

The first line contains a single integer n (2≤n≤105) — the number of heart-shaped cards.

The second line contains n−1 integers p2,p3,…,pn (1≤pi<i) describing which card that each card hangs onto.

Output

Print a single integer — the maximum length of the longest non-decreasing subsequence of s at the end if Pak Chanek does all the steps optimally.

Examples

input

Copy

6
1 2 1 4 2

output

Copy

4

input

Copy

2
1

output

Copy

2

题意： 给定一个树，我们的目标是构造树的点权。每次操作可以选择一个叶子节点，并且摘掉这个叶子节点并且获得叶子节点的权值。每次摘叶子的时候，如果你摘的叶子的权值比他的父节点小，那么父节点的权值就会变成这个叶子的权值，一直进行到树被摘完。最终我们会获得摘叶子的序列，起价值是该序列的最长非下降子序列。请合理的构造树上的权值，最大化操作后的价值。

代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;

#define all(x)      (x).begin(),(x).end()
#define X           first
#define Y           second
#define sep         ' '
#define endl        '\n'
#define SZ(x)       ll(x.size())

const ll MAXN = 1e6 + 10;
const ll LOG = 22;
const ll INF = 8e18;
const ll MOD = 1e9 + 7; //998244353; //1e9 + 9;

int pow_mod(int a1, int n1, int m1)
{
    if(n1 == 0) return 1;
    int x = pow_mod(a1, n1/2, m1);
    long long ans = (long long)x * x % m1;
    if(n1 % 2 == 1) ans = ans *a1 % m1;
    return (int)ans;
}
int n , ans , P[MAXN] , dp[MAXN] , mx[MAXN];
vector<int> adj[MAXN];

int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    cin >> n;
    for(int i = 2 ; i <= n ; i++){
        cin >> P[i];
        adj[P[i]].push_back(i);
    }
    for(int i = n ; i >= 1 ; i--){
        for(int j : adj[i]){
            dp[i] += dp[j];
            mx[i] = max(mx[i] , mx[j]);
        }
        mx[i]++;
        dp[i] = max(dp[i] , mx[i]);
    }
    cout << dp[1] << endl;

    return 0;
}