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.

Examples

Input

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


Output

8


Input

2 3
2 1
10 -7


Output

13


Input

3 3
3 1 2
-1000 -2000 -3000


Output

-1000


Input

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


Output

29507023469
#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <iomanip>
#include <string.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()

typedef long long lint;

using namespace std;

int main(){
  lint N,K;
  cin>>N>>K;
  lint P[N],C[N];
  REP(i,N){
    cin>>P[i];
    P[i]--;
  }
  REP(i,N)cin>>C[i];
  lint ans=-1e18;
  REP(i,N){
    int p=i;
    lint k=K;
    lint temp=0;
    REP(j,k){
      p=P[p];
      temp+=C[p];
      if(p==i){
        int c=max(0LL,k/(j+1)-2);
        temp+=temp*c;
        k-=(j+1)*c;
        cerr<<i<<" "<<j<<" "<<k<<" "<<c<<endl;
      }
      ans=max(ans,temp);
    }
  }
  cout<<ans<<endl;
  return 0;
}
