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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;
use std::cmp::max;

fn main() {
    input! {
        N: usize,
        K: i64,
        mut Ps: [usize; N],
        Cs: [i64; N],
    }

    // Psの値は1以上なので、0オリジンにする
    for i in 0..N {
        Ps[i] -= 1;
    }

    // ループがある。
    // O(N)で開始点ごとの1周の周期とポイントを調べて、
    // 次に、O(N)で各点スタートのときの点を調べて最大のものを調べる?

    // periods[i] : iを開始点とした時の周期
    let mut periods: Vec<i64> = vec![0i64; N];

    // points[i] : iを開始点とした時の1周期で得られるポイント
    let mut points: Vec<i64> = vec![0i64; N];

    // 周期とポイントを調べる
    for i in 0..N {
        // もう確認済みならばスキップ (後述)
        if periods[i] != 0 {
            continue;
        }


        // whileでうまく回すために、あらかじめ1ムーブしておく
        let mut current = Ps[i];
        let mut period = 1;
        let mut point = Cs[current];

        // マスとマスjが同じループの中にあるならば、periods[i]==points[j]が成り立つ (periodsも同様)
        // よって、通ったマスを記憶しておくことで、ループ回数を減らすことができる
        let mut paths: Vec<usize> = Vec::new();
        paths.push(i);
        paths.push(current);

        while current != i {
            current = Ps[current];
            period += 1;
            point += Cs[current];
        }

        for p in paths.iter() {
            periods[*p] = period;
            points[*p] = point;
        }
    }

    for i in 0..N {
        dbg!(periods[i]);
        dbg!(points[i]);
    }

    let mut ans = std::i64::MIN;
    // 実際に巡ってみよう
    for i in 0..N {
        // まず1回は進むのがルール
        let mut current = Ps[i];
        let mut point: i64 = Cs[current];
        ans = max(ans, point);
        let mut rest = K - 1;

        // まず、周期でざっとやる
        // ただし、損するならやらない
        if points[current] > 0 {
            let cnt: i64 = rest / periods[current];
            point += points[current] * cnt;
            ans = max(ans, point);
            rest -= periods[current] * cnt;
        } else {
            // 1周したら損するので、1周以下を見ればよい
            rest = periods[current];
        }

        // 1歩ずつ進んで、点を更新していく
        // いつでもやめられるので、最高点をマークし続ければOK
        // restは5000で割った余りになっているので、5000 * 5000のオーダーとなる。
        // ギリギリ間に合うかな?
        for _k in 0..rest {
            current = Ps[current];
            point += Cs[current];
            ans = max(ans, point);
        }
    }

    println!("{}", ans);
}