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)

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


#define typeof(x) __typeof__(x)
#define bit(x,i) (x&(1<<i))  //select the bit of position i of x
#define lowbit(x) ((x)&((x)^((x)-1))) //get the lowest bit of x
#define hBit(msb,n) asm("bsrl %1,%0" : "=r"(msb) : "r"(n)) //get the highest bit of x, maybe the fastest
#define max(a,b) (a<b?b:a)
//#define abs(x) (x<0?(-x):x) // big bug here if "-x" is not surrounded by "()"
#define IN(i,l,r) (l<i&&i<r) //the next for are for checking bound
#define LINR(i,l,r) (l<=i&&i<=r)
#define LIN(i,l,r) (l<=i&&i<r)
#define INR(i,l,r) (l<i&&i<=r)
#define F(i,L,R) for (int i = L; i < R; i++) //next four are for "for loops"
#define FIN(i,L,R) for (int i = L; i <= R; i++)
#define FF(i,L,R) for (int i = L; i > R; i--)
#define FFIN(i,L,R) for (int i = L; i >= R; i--)
#define getI(a) scanf("%d", &a) //next three are handy ways to get ints, it's also force you to use '&' sign
#define getII(a,b) scanf("%d%d", &a, &b)
#define getIII(a,b,c) scanf("%d%d%d", &a, &b, &c)
#define getLL(a) scanf("%lld", &a)
#define g_s(n) int (n); scanf("%d",&(n)) //handy if the input is right after the definition of a variable
#define g_s2(n,m) int (n),(m); scanf("%d %d",&(n),&(m))
#define g_s3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k))
#define TESTS g_s(testow);while(testow--) //for multilple cases problems
#define whileZ int T; getI(T); while(T--) // the same as above
#define getS(x) scanf("%s", x) //get a char* string
#define MEM(a,x) memset(a,x,sizeof(a)) //set elements of array to some value
#define char2Int(c) (c-'0')
#define lastEle(vec) vec[vec.size()-1]
#define SZ(x) ((int)((x).size()))
#define REMAX(a,b) (a)=max((a),(b)) // set a to the maximum of a and b
#define REMIN(a,b) (a)=min((a),(b));
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) // traverse an STL data structure
#define ALL(c) (c).begin(),(c).end() //handy for function like "sort()"
#define PRESENT(c,x) ((c).find(x) != (c).end())
#define CPRESENT(c,x) (find(ALL(c),x) != (c).end())
#define ll long long //data types used often, but you don't want to type them time by time
#define ull unsigned long long
#define ui unsigned int
#define us unsigned short
#define IOS ios_base::sync_with_stdio(0); //to synchronize the input of cin and scanf
#define INF 1001001001
#define PI 3.1415926535897932384626
//for map, pair
#define mp make_pair
#define fi first
#define se second
// for debug
inline void pisz(int n) { printf("%d\n",n); }
#define DBG(vari) cerr<<#vari<<" = "<<(vari)<<endl;
#define printA(a,L,R) FIN(jk,L,R) cout << a[jk] << (jk==R?'\n':' ')
#define printV(a) printA(a,0,a.size()-1)
#define MAXN 10000
#define sf scanf
#define pf printf
//for vectors
#define pb push_back
typedef int elem_t;
typedef vector<int> vi;
typedef vector<string> vstr;
typedef vector<set<int> > vset;
typedef vector<vi> vvi;
typedef vector<bool> vb;
typedef vector<vector<bool> > vvb;
typedef pair<int,int> ii;
typedef vector<ii> vii;
// directions
const int dr[4] = {-1,0,1,0};
const int dc[4] = {0,-1,0,1};
const int drr[8] = {-1,-1,0,1,1,1,0,-1};
const int dcc[8] = {0,-1,-1,-1,0,1,1,1};
template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";}
template<typename T> ostream& operator<<(ostream &s,vector<T> t){F(i,0,SZ(t))s<<t[i]<<" ";return s; }

bool ckmin(int& a, int b){ return b < a ? a = b, true : false; }
bool ckmax(int& a, int b){ return b > a ? a = b, true : false; }

class UnionFind{
    private: vi rank,p;
    public:
        UnionFind(int n){
            F(i,0,n)p.pb(i);
            rank.assign(n,1);
        }

        int findSet(int s){
            return (p[s]==s) ? s : (p[s]=findSet(p[s]));
        }

        bool isSameSet(int i,int j){
            return findSet(i)==findSet(j);
        }

        void unionSet(int i,int j){
            if(isSameSet(i,j))return;

            int x=findSet(i),y=findSet(j);
            if(rank[x]>rank[y]){
                p[y]=x;
                rank[x]+=rank[y];
            }
            else {
                p[x]=y;
                //if(rank[x]==rank[y])rank[y]++;
                rank[y] += rank[x];
            }

        }

        int sz(int x){
            int px=findSet(x);
            return rank[px];
        }
};


double x_y(int x1,int y1,int x2,int y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}



int main(){
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr);
       
   // g_s2(n,k);
    ll n,k;
     getLL(n);
    getLL(k);
    ll p[n+1],c[n+1];
   

    F(i,1,n+1)getLL(p[i]);
    F(i,1,n+1)getLL(c[i]);

    ll ans = -1e18;
  //  cout<<ans<<endl;

    F(i,1,n+1){
        ll vis[n+1],moves=0,nwk=k,val=i;
        ll nwdt[n+5];
        MEM(nwdt,0);
        MEM(vis,0);
        ll cur=0;

        while(nwk && !vis[val]){
            cur += c[p[val]];
            ans = max(ans,cur);
            vis[val]=1;
            val = p[val];
            nwdt[++moves]=cur;
            //moves++;
            nwk--;
        }
       // cout<<nwdt[0]<<endl;

        if(!nwk)continue;
        ll iv=0;
        iv= (ll)((ll)k/(ll)moves)*nwdt[moves];

        F(j,1,k%moves+1){
            //if(vis[j]){
                //ll iv= ((k-j)/moves+1)*nwdt[j];
                
               // if((k%moves)>=j)iv += nwdt[j];

             //   cout<<j<<"   "<<iv<<"   "<<i<<"  mv  "<<moves<<" tot  "<<nwdt[moves]<<endl;
                ans = max(ans,iv+nwdt[j]);
            
        }

    }

    cout<<ans<<endl;

    return 0;
}