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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"container/heap"
	"fmt"
	"os"
	"sort"
	"strconv"
)

const (
	initialBufSize = 1e4
	maxBufSize     = 1e6 + 7
)

var buf []byte = make([]byte, maxBufSize)
var sc = bufio.NewScanner(os.Stdin)

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buf, maxBufSize)
}

func main() {
	N, K := nextInt(), nextInt()
	P := make([]int, N)
	C := make([]int, N)
	for i := 0; i < N; i++ {
		P[i] = nextInt() - 1
	}
	for i := 0; i < N; i++ {
		C[i] = nextInt()
	}
	// fmt.Println(P)
	cycle := make([]int, N)
	for i := 0; i < N; i++ {
		check := make(map[int]bool)
		location := i
		for j := 0; j < N; j++ {
			if _, e := check[P[location]]; e {
				cycle[i] = j
				break
			}
			check[P[location]] = true
			location = P[location]
		}
	}
	// fmt.Println(cycle)
	memo := make([][]int, N)
	max := -int(1e9) - 7
	for i := 0; i < N; i++ {
		l := cycle[i]
		if l == 0 {
			l = N
		}
		memo[i] = make([]int, l+1)
		location := i
		for j := 0; j < l; j++ {
			// if j >= K {
			// 	break
			// }
			memo[i][j+1] = memo[i][j] + C[location]
			location = P[location]
			max = MaxOf(max, memo[i][j+1])
		}
	}
	// fmt.Println(memo)
	if max <= 0 {
		fmt.Println(max)
		return
	}
	for i := 0; i < N; i++ {
		l := len(memo[i]) - 1
		e := (K - i - 1) / l
		summax := e * memo[i][l]
		start := summax
		for j := 0; j < K%l; j++ {
			// fmt.Println(summax)
			summax = MaxOf(summax, start+memo[i][j+1])
		}
		max = MaxOf(max, summax)
	}
	fmt.Println(max)
}

func nextLine() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func abs(a int) int {
	if a > 0 {
		return a
	} else {
		return -a
	}
}

func pow(p, q int) int {
	res := p
	if q == 0 {
		return 1
	}
	for i := 0; i < q-1; i++ {
		res *= p
	}
	return res
}

func ceil(a, b int) int {
	return (a + (b - 1)) / b
}

func MinOf(vars ...int) int {
	min := vars[0]
	for _, i := range vars {
		if min > i {
			min = i
		}
	}
	return min
}

func MaxOf(vars ...int) int {
	max := vars[0]
	for _, i := range vars {
		if max < i {
			max = i
		}
	}
	return max
}

func gcdof2numbers(a int, b int) int {
	if b == 0 {
		return a
	}
	return gcdof2numbers(b, a%b)
}

func lcmof2numbers(a int, b int) int {
	return a / gcdof2numbers(a, b) * b
}

// マイナスの場合は考慮していない
func fuctorial(a int) int {
	if a == 1 || a == 0 {
		return 1
	} else {
		return fuctorial(a-1) * a
	}
}

func NextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}

// 素因数分解したmapを返す
func primeFuctorize(n int) map[int]int {
	pf := map[int]int{}
	for i := 2; i*i <= n; i++ {
		for n%i == 0 {
			pf[i]++
			n /= i
		}
	}
	if n != 1 {
		pf[n]++
	}
	return pf
}

// ここから拝借
// https://shoman.hatenablog.com/entry/2020/02/25/185456
// Stackは[]intのエイリアス
type Stack []int

// Push adds an element
func (s *Stack) Push(v int) {
	*s = append(*s, v)
}

// Pop removes the top element and return it
func (s *Stack) Pop() (int, error) {
	if s.Empty() {
		return 0, fmt.Errorf("stack is empty")
	}

	v := (*s)[len(*s)-1]
	*s = (*s)[:len(*s)-1]
	return v, nil
}

// Peek returns the top value
func (s *Stack) Peek() (int, error) {
	if s.Empty() {
		return 0, fmt.Errorf("stack is empty")
	}

	return (*s)[len(*s)-1], nil
}

// Size returns the length of stack
func (s *Stack) Size() int {
	return len(*s)
}

// Empty returns true when stack is empty
func (s *Stack) Empty() bool {
	return s.Size() == 0
}

// NewStack generates stack
func NewStack() *Stack {
	s := new(Stack)
	return s
}

// golangの公式サンプルより
// https://xn--go-hh0g6u.com/pkg/container/heap/#example__intHeap
// IntHeap は，整数の最小ヒープです。
type IH []int

func (h IH) Len() int           { return len(h) }
func (h IH) Less(i, j int) bool { return h[i] < h[j] }
func (h IH) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IH) Push(x interface{}) {
	// Push と Pop はポインタレシーバを使っています。
	// なぜなら，スライスの中身だけでなく，スライスの長さも変更するからです。
	*h = append(*h, x.(int))
}

func (h *IH) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

// golangの公式サンプルより
// https://xn--go-hh0g6u.com/pkg/container/heap/#example__priorityQueue

// Item は，優先キューで管理する項目です。
type Item struct {
	value    string // 値。任意です。
	priority int    // キューにおける優先度
	// index は heap.Interface メソッドで更新されます。
	index int // ヒープにおけるインデックス
}

// PriorityQueue は heap.Interface を実装し，Item のリストを保持します。
type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }

func (pq PriorityQueue) Less(i, j int) bool {
	// Pop が最小ではなく最大の優先度を持つ項目を返して欲しいので，ここでは > を使っています。
	return pq[i].priority > pq[j].priority
}

func (pq PriorityQueue) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *PriorityQueue) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Item)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *PriorityQueue) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // メモリリークを避ける
	item.index = -1 // 安全のため
	*pq = old[0 : n-1]
	return item
}

// update はキューの Item の優先度と値を更新します。
func (pq *PriorityQueue) update(item *Item, value string, priority int) {
	item.value = value
	item.priority = priority
	heap.Fix(pq, item.index)
}