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<bits/stdc++.h>
#define ALL(c) begin(c), end(c)
using namespace std;
using ll = long long;
using ld = long double;

const int N = 1e6+6;


int main(){
	//freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout);
	ios::sync_with_stdio(0);cin.tie(0);//cout.precision(12);cout<<fixed;
	
	int n, d;
	
	cin>>n>>d;
	
	vector<int> p(n);
	for(int i=0;i<n;++i){
		cin>>p[i];
		--p[i];
	}
	
	vector<int> c(n);
	for(auto &_ : c) cin>>_;
	
	ll ans = -1e18;
	for(int i=0;i<n;++i){
		vector<bool> u(n);
		ll sum = 0;
		int len = 0;
		for(int k=i;!u[k];k=p[k]){
			u[k] = 1;
			++len;
			sum+=c[k];
		}
		
		
		ll pref = 0;
		for(int h=1,k=i; h<=len; ++h,k=p[k]){
			pref+=c[k];
			if(h>d) break;
			
			int x = (d - h) / len;
			
			ans = max(ans, sum * x + pref);
			ans = max(ans, pref);
		}
	}
	
	cout<<ans<<endl;
	
	return 0;
}
