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<algorithm>     //sort,二分探索,など
#include<bitset>        //固定長bit集合
#include<cmath>         //pow,logなど
#include<complex>       //複素数
#include<deque>         //両端アクセスのキュー
#include<fstream>       //ファイルストリーム（標準入力変更用）
#include<functional>    //sortのgreater
#include<iomanip>       //setprecision(浮動小数点の出力の誤差)
#include<iostream>      //入出力
#include<iterator>      //集合演算(積集合,和集合,差集合など)
#include<map>           //map(辞書)
#include<numeric>       //iota(整数列の生成),gcdとlcm(c++17)
#include<queue>         //キュー
#include<set>           //集合
#include<stack>         //スタック
#include<string>        //文字列
#include<unordered_map> //イテレータあるけど順序保持しないmap
#include<unordered_set> //イテレータあるけど順序保持しないset
#include<utility>       //pair
#include<vector>        //可変長配列

//名前
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef map<string, int> msi;
typedef map<string, ll> msll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pllll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvll;
typedef vector<vector<string>> vvs;
typedef vector<vector<bool>> vvb;

//定数
const ll MOD = 1000000007;
const ll INF = 1000000000000000000;

//マクロ
#define rep(i,n) for(int i=0;i<n;i++)
#define reps(i,s,e) for(int i=s;i<e;i++)
#define repse(i,s,e) for(int i=s;i<=e;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define in(x1) cin >> x1
#define in2(x1, x2) cin >> x1 >> x2
#define in3(x1, x2, x3) cin >> x1 >> x2 >> x3
#define inN(x, N) rep(i, N) in(x[i])
#define outl(x) cout << x << endl
#define out2l(x, y) cout << x << " " << y << endl 

//よく使う関数
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }


class UnionFind
{
	//par[i]：ノードiの親
	//count[i]：iが親の時、iが属するグループのノード数
	vector<int> par;
	vector<int> count;

	int m_groupCount = 0;

public:
	UnionFind(int N)
	{
		//1始まりのときに必要
		//par.push_back(0);
		//count.push_back(0);

		m_groupCount = N;

		rep(i, N)
		{
			//最初は全てが根であるとして初期化
			par.push_back(i);
			count.push_back(1);
		}
	}

	//データxが属する木の根を再帰で得る：root(x) = {xの木の根}
	int root(int x)
	{
		if (par[x] == x) return x;

		//ついでに親を張り替え
		return par[x] = root(par[x]);
	}

	//xとyの木を併合
	void unite(int x, int y)
	{
		int rx = root(x);
		int ry = root(y);
		if (rx == ry) return;
		//xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
		par[rx] = ry;

		//要素数の変更
		count[ry] += count[rx];
		count[rx] = 0;

		//木の併合によってグループが1つ減る
		m_groupCount--;
	}

	// 2つのデータx, yが属する木が同じならtrueを返す
	bool same(int x, int y)
	{
		return root(x) == root(y);
	}

	//指定したデータxが属するグループの要素数を返す
	int size(int x)
	{
		return count[root(x)];
	}

	//UnionFindのグループの個数を返す
	int groupCount()
	{
		return m_groupCount;
	}

	//UnionFindのコピー処理
	UnionFind copy()
	{
		UnionFind uf(par.size());

		//元のUnionFindの親情報を再現
		rep(i, par.size())
		{
			uf.unite(i, this->root(i));
		}

		return uf;
	}
};


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

    //標準入力をファイルに変更
    //std::ifstream in("input.txt");
    //std::cin.rdbuf(in.rdbuf());

    int N, K;
    in2(N, K);

    vi P(N);
    inN(P, N);
	rep(i, N) P[i]--;

    vll C(N);
    inN(C, N);

	UnionFind uf(N);
	rep(i, N) uf.unite(i, P[i]);

	ll ans = -INF;

	rep(i, N)
	{
		int count = K;
		ll tmp = 0;
		ll sum = 0;
		vll score;
		int pos = i;

		do
		{
			pos = P[pos];
			sum += C[pos];
			score.push_back(C[pos]);
		} while (pos != i);

		if (sum > 0)
		{
			tmp += sum * (count / uf.size(i));
			count %= uf.size(i);
		}

		rep(i, score.size() - 1) score[i + 1] += score[i];
		ll y = score[0];
		rep(i, min(count, (int)score.size())) chmax(y, score[i]);
		tmp += y;

		chmax(ans, tmp);
	}

	outl(ans);

    return 0;
}