Problem:
Score : 400 points

Problem Statement

Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.

Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):

In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.

Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)

Constraints

2 \leq N \leq 5000

1 \leq K \leq 10^9

1 \leq P_i \leq N

P_i \neq i

P_1, P_2, \cdots, P_N are all different.

-10^9 \leq C_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N

Output

Print the maximum possible score at the end of the game.

Sample Input 1

5 2
2 4 5 1 3
3 4 -10 -8 8

Sample Output 1

8

When we start at some square of our choice and make at most two moves, we have the following options:

If we start at Square 1, making one move sends the piece to Square 2, after which the score is 4. Making another move sends the piece to Square 4, after which the score is 4 + (-8) = -4.

If we start at Square 2, making one move sends the piece to Square 4, after which the score is -8. Making another move sends the piece to Square 1, after which the score is -8 + 3 = -5.

If we start at Square 3, making one move sends the piece to Square 5, after which the score is 8. Making another move sends the piece to Square 3, after which the score is 8 + (-10) = -2.

If we start at Square 4, making one move sends the piece to Square 1, after which the score is 3. Making another move sends the piece to Square 2, after which the score is 3 + 4 = 7.

If we start at Square 5, making one move sends the piece to Square 3, after which the score is -10. Making another move sends the piece to Square 5, after which the score is -10 + 8 = -2.

The maximum score achieved is 8.

Sample Input 2

2 3
2 1
10 -7

Sample Output 2

13

Sample Input 3

3 3
3 1 2
-1000 -2000 -3000

Sample Output 3

-1000

We have to make at least one move.

Sample Input 4

10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719

Sample Output 4

29507023469

The absolute value of the answer may be enormous.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fna")
#pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define pb push_back
#define mp make_pair
#define endl "\n"
#define int ll
#define vi vector<int>
#define vb vector<bool>
#define vvb vector<vb >
#define pii pair<int,int>
#define ss second
#define ff first
#define vpii vector<pii>
#define vvi vector<vi >
#define vs vector<string>
#define vvs vector<vs >
#define pqi priority_queue <int>
#define minpqi priority_queue <int, vector<int>, greater<int> >
#define all(x) x.begin(),x.end()
#define mii map<int,int>
#define for0(i,n) for(ll i=0;i<n;i++)
#define for1(i,n) for(ll i=1;i<=n;i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define per1(i,n) for(ll i=n;i>0;i--)
#define repeat(i,start,n) for(ll i=start;i<n;i++)
#define inp(arr,n) ll arr[n];rep(i,n){ cin>>arr[i];}
#define inp1(arr,n) ll arr[n+1];rep1(i,n){ cin>>arr[i];}
#define inp2d(arr,n,m) ll arr[n][m];rep(i,n)rep(j,m)cin>>arr[i][j];
#define inpg(adj,m) rep(i,m){int a,b;cin>>a>>b;adj[a].pb(b);adj[b].pb(a);}
#define print(a,n) for(ll i=0;i<n;i++){ cout<<a[i]<<" ";}
#define print1(a,n) for(ll i=1;i<=n;i++){ cout<<a[i]<<endl;}
#define mod 1000000007
#define maxx 1000000000000000000
#define PI 3.141592653589793238462643383279
#define mmax(a,b,c) max(a,max(b,c))
#define mmin(a,b,c) min(a,min(b,c))
#define init(arr,a) memset(arr,a,sizeof(arr))
#define lb lower_bound
#define ub upper_bound
#define er equal_range
#define maxe *max_element
#define mine *min_element
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b){
	if(a.ss==b.ss)
		return a.ff<b.ff;
	return (a.second<b.second);
}
string convert_to_bit(int a,int bit){
	string s;
	while(a>0){
		s+=(a%2)+48;
		a/=2;
	}
	while(s.size()<bit)
		s+='0';
	reverse(s.begin(),s.end());
	return s;
}
int to_int(string s)
{
	int ans=0;
	string temp;
	int i=0;
	while(s[i]=='0')
	{
		i++;
	}
	if(i==s.size())
		return 0;
	temp.assign(s,i,s.size());
	int mul=1;
	for(i=temp.size()-1;i>=0;i--)
	{
		int a=temp[i]-'0';
		ans+=mul*a;
		mul*=10;
	}
	return ans;

}
string to_string(int n)
{
	string ans="";
	if(n==0)
		return "0";
	while(n>0)
	{
		int a=n%10;
		n/=10;
		char temp=a+'0';
		ans+=temp;
	}
	reverse(all(ans));
	return ans;
}
int bin_to_dec(int n)
{
	int num = n;
	int dec_value = 0;
	int base = 1;
	int temp = num;
	while (temp) {
		int last_digit = temp % 10;
		temp = temp / 10;
		dec_value += last_digit * base;
		base = base * 2;
	}
	return dec_value;
}
int binstr_to_dec(string s)
{
	int ans=0;
	for0(i,s.size())
		if(s[i]=='1')
			ans+=pow(2,s.size()-i-1);
	return ans;
}
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
void faltu()
{
	int a=2;
	while(a>0)a--;
}
int nCr(ll n, ll r)
{
	ll res = 1;
	if ( r > n - r )
		r = n-r;
	for (ll i = 0; i < r; ++i)
	{
		res *= (n - i);	//general term (n-x)/(1+x)
		res /= (i + 1);
	}
	return res;
}
//------------------------------------Code Starts here------------------------------------//

void solve()
{

}
main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n,k;
	cin>>n>>k;
	int p[n+1],c[n+1];
	for1(i,n)cin>>p[i];
	for1(i,n)cin>>c[i];
	int cnt=0;vi v;
	int maxi=-1,pos=0;
	for1(i,n)
	{
		if(c[i]>=0)pos++;
	}
	if(pos==0)
	{
		maxi=-1000000000000;
		for1(i,n)
		maxi=max(maxi,c[i]);
cout<<maxi;
	}

	else
  {for1(i,n)
  {	int z=i,sum=0;int zz=1;
  int l=min(k,(int)10000);
			while(zz<=l&&sum>=0)
			{
				sum+=c[p[z]];
				maxi=max(sum,maxi);
				z=p[z];
				zz++;
				//if(c[p[z]]>=0)pos++;
			}
  }

	cout<<maxi<<endl;
  }



}