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(unused_imports)]

// language: Rust(1.42.0)
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"
// My Library Repositry is at "https://github.com/Loptall/sfcpl"

/*
    水コーダーになれてうれしいです
*/

// from std...
use std::cmp::{
    max, min, Ordering,
    Ordering::{Equal, Greater, Less},
    Reverse,
};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::convert::TryInto;
use std::fmt;
use std::mem::swap;
use std::num::{NonZeroU32, ParseIntError};
use std::ops::{
    Add, AddAssign, BitAnd, BitAndAssign, BitOr, BitOrAssign, BitXor, BitXorAssign, Div, DivAssign,
    Index, IndexMut, Mul, MulAssign, Neg, Rem, RemAssign, Shl, ShlAssign, Shr, ShrAssign, Sub,
    SubAssign,
};
use std::process::exit;
use std::{f32, f64, i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

// Dep' crates are...
use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_complex::Complex;
use num_integer::{binomial, gcd, lcm, multinomial, Integer};
use num_rational::Rational;
use num_traits::{
    clamp, one, pow, zero, Num, NumAssignOps, NumOps, One, Pow, Signed, Unsigned, Zero,
};
use permutohedron::Heap;
use proconio::{
    derive_readable, fastout, input, is_stdin_empty,
    marker::{Bytes, Chars, Isize1, Usize1},
};
use rand::random;

pub trait Visualize {
    fn visualize(&self, split: &str);
    fn continuous(&self) {
        self.visualize("");
    }
    fn spaces(&self) {
        self.visualize(" ");
    }
    fn lines(&self) {
        self.visualize("\n");
    }
}
macro_rules! impl_vis_for_sized {
    ($($t:ty),+) => {
        $(
            impl Visualize for $t {
                fn visualize(&self, _split: &str) {
                    print!("{}", self);
                }
            }
        )+
    };
}
impl_vis_for_sized! {
    usize, u8, u16, u32, u64, u128,
    isize, i8, i16, i32, i64, i128,
    f64, f32,
    String, &str, char
}
impl<T: fmt::Display> Visualize for [T] {
    fn visualize(&self, split: &str) {
        print!("{}", self.iter().join(split));
    }
}
#[macro_export]
macro_rules! vis {
    // end
    () => {
        println!();
    };

    // last element + trailing pattern
    ($last:expr ;) => {
        $last.lines();
        vis!()
    };
    ($last:expr =>) => {
        $last.continuous();
        vis!();
    };
    ($last:expr $(,)?) => {
        $last.spaces();
        vis!();
    };

    // get first element and pass rest
    ($first:expr; $($rest:tt)*) => {
        $first.lines();
        println!();
        vis!($($rest)*);
    };
    ($first:expr => $($rest:tt)*) => {
        $first.continuous();
        vis!($($rest)*);
    };
    ($first:expr, $($rest:tt)*) => {
        $first.spaces();
        print!(" ");
        vis!($($rest)*);
    };
}
pub const MOD10E9_7: usize = 1000000007; // 10 ^ 9 + 7
pub const MOD99_: usize = 998244353;
pub const MAX: usize = std::usize::MAX; // = 2 ^ 64 - 1 = 18446744073709551615 ≈ 1.8 * 10 ^ 19
pub const INF: usize = 2000000000000000000; // MAX / 9 < 2 * 10e18 < MAX / 10
pub const FNI: i64 = -2000000000000000000; // == -(INF as i64)
pub const PI: f64 = f64::consts::PI; // 3.141592653589793 -- 10 ^ -15
pub const ASCII_A_LARGE: u8 = 65;
pub const ASCII_A_SMALL: u8 = 97;
pub const ASCII_0: u8 = 48;
pub const ADJ4: &[(isize, isize); 4] = &[(1, 0), (0, 1), (-1, 0), (0, -1)];
pub const ADJ8: &[(isize, isize); 8] = &[
    (1, 0),
    (1, 1),
    (0, 1),
    (-1, 1),
    (-1, 0),
    (-1, -1),
    (0, -1),
    (1, -1),
];

#[derive(Debug, Clone)]
pub struct UnionFind {
    parent: Vec<usize>,
    size: Vec<usize>,
}
impl UnionFind {
    pub fn new(n: usize) -> Self {
        Self {
            parent: (0..n).collect::<Vec<usize>>(),
            size: vec![1; n],
        }
    }
    pub fn len(&self) -> usize {
        self.parent.len()
    }
    /// i が属する集合の親のインデックス
    pub fn find(&mut self, mut i: usize) -> usize {
        while self.parent[i] != i {
            self.parent[i] = self.find(self.parent[i]);
            i = self.parent[self.parent[i]];
        }
        i
    }
    /// a と b を繋ぐ
    pub fn unite(&mut self, a: usize, b: usize) -> bool {
        let mut a = self.find(a);
        let mut b = self.find(b);
        if a == b {
            return false;
        }
        if self.size[a] < self.size[b] {
            std::mem::swap(&mut a, &mut b);
        }
        self.size[a] += self.size[b];
        self.parent[b] = a;
        true
    }
    /// i が属する集合の要素数
    pub fn count(&mut self, i: usize) -> usize {
        let p = self.find(i);
        self.size[p]
    }
    /// a と b が同一集合に属するか
    pub fn joint(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }
    /// i が属する集合の要素を列挙する
    pub fn get_group(&mut self, i: usize) -> HashSet<usize> {
        let p = self.find(i);
        (0..self.len()).filter(|x| self.find(*x) == p).collect()
    }
}

// code...
// #[fastout]
fn main() {
    input! {
        n: usize, k: usize,
        p: [Usize1; n],
        c: [i64; n]
    }

    let mut uf = UnionFind::new(n);
    for (i, &j) in p.iter().enumerate() {
        uf.unite(i, j);
    }

    let mut ans = FNI;
    for start in 0..n {
        let cycle = uf.count(start);

        let mut total = 0;
        let mut path = Vec::new();
        let mut cur = start;
        for _ in 0..cycle {
            cur = p[cur];
            total += c[cur];
            path.push(cur);
        }

        // assert_eq!(path.len(), cycle);
        // dbg!(&path);

        let mut acc = 0i64;
        for i in 0..cycle {
            acc += c[path[i]];
            if i + 1 > k {
                break;
            }
            let mut e = 0;
            if total > 0 {
                e = (k - i - 1) / cycle;
            }

            ans = max(ans, acc + total * e as i64);
        }
    }

    vis!(ans);
}