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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	static int[][] map;
	static int[][] directions8= {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
	static int[][] directions4= {{-1,0},{1,0},{0,-1},{0,1}};
	static long ans;
	static int[] parent;//union-find用
	static int[] rank;//union-find用

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();

		int[] p = new int[N+1];
		long[] c = new long[N+1];
		ans = 0;


		for(int i = 1;i<=N ; i++) {
			p[i] = sc.nextInt();
		}

		long cSum = 0;
		long maxC = Long.MIN_VALUE;

		for(int i = 1;i<=N ; i++) {
			c[i] = sc.nextLong();
			cSum+=c[i];
			maxC = Long.max(maxC, c[i]);
		}


		if(cSum>0) {
			ans+=(K/N)*cSum;
		}

		K%=N;

		int len = 0;
		long max = 0;

		for(int start = 1;start<=N ; start++) {
			long tmpSum = 0;
			int l = start;
			for(int i = 0;i<K ; i++) {
				tmpSum+=c[l];
				if(max<tmpSum) {
					max=tmpSum;
				}
				l=p[l];
			}

		}
		ans+=max;

		if(ans == 0) {
			ans = maxC;
		}

		System.out.println(ans);
	}

	//素数判定
	public static boolean isPrime(long num) {
		if (num < 2) return false;
		else if (num == 2) return true;
		else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

		double sqrtNum = Math.sqrt(num);
		for (int i = 3; i <= sqrtNum; i += 2)
		{
			if (num % i == 0)
			{
				// 素数ではない
				return false;
			}
		}

		// 素数である
		return true;
	}

	//BFS用に二つの配列を足し算する
	static int[] addArrayElms(int[] a, int[] b) {
		int[] c = new int[a.length];
		for(int i = 0; i < a.length; i++) {
			c[i] = a[i] + b[i];
		}
		return c;
	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
		return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
		return b == 0 ? a: gcd(b, a % b);
	}

	//Union-Find用の配列を要素数nで初期化
	static void ufInit(int n){
		parent = new int[n];
		rank = new int[n];
	}

	//Union-Findの要素の最上位の親（グループ名）を返す
	static int ufFind(int a) {
		if(parent[a]==a) {
			return a;
		}else {
			return ufFind(parent[a]);
		}
	}

	//Union-Find木を連結する
	static void ufUnite(int a, int b) {
		a = parent[a];
		b = parent[b];

		if(a==b) {
			return;
		}
		if(rank[a] > rank[b]){
			parent[b]=a;
		}else {
			parent[a]=b;
			if(rank[a]==rank[b]) {
				rank[b]++;
			}
		}
	}

	static boolean ufSame(int a, int b) {
		return ufFind(a)==ufFind(b);
	}
}