Problem:
Score : 400 points

Problem Statement

At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:

The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)

Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):

R points for winning with Rock;

S points for winning with Scissors;

P points for winning with Paper.

However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)

Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.

The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is r, the machine will play Rock in the i-th round. Similarly, p and s stand for Paper and Scissors, respectively.

What is the maximum total score earned in the game by adequately choosing the hand to play in each round?

Notes

In this problem, Rock Paper Scissors can be thought of as a two-player game, in which each player simultaneously forms Rock, Paper, or Scissors with a hand.

If a player chooses Rock and the other chooses Scissors, the player choosing Rock wins;

if a player chooses Scissors and the other chooses Paper, the player choosing Scissors wins;

if a player chooses Paper and the other chooses Rock, the player choosing Paper wins;

if both players play the same hand, it is a draw.

Constraints

2 \leq N \leq 10^5

1 \leq K \leq N-1

1 \leq R,S,P \leq 10^4

N,K,R,S, and P are all integers.

|T| = N

T consists of r, p, and s.

Input

Input is given from Standard Input in the following format:

N K
R S P
T

Output

Print the maximum total score earned in the game.

Sample Input 1

5 2
8 7 6
rsrpr

Sample Output 1

27

The machine will play {Rock, Scissors, Rock, Paper, Rock}.

We can, for example, play {Paper, Rock, Rock, Scissors, Paper} against it to earn 27 points.
We cannot earn more points, so the answer is 27.

Sample Input 2

7 1
100 10 1
ssssppr

Sample Output 2

211

Sample Input 3

30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr

Sample Output 3

4996

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.function.BiFunction;
import java.util.function.Consumer;
import java.util.function.Supplier;

public class Main {
    final static long MOD07 = 1_000_000_007;
    final static long MOD09 = 1_000_000_009;

    public static void main(String[] args) {
        final FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        int N = fs.nextInt();
        int K = fs.nextInt();
        long R = fs.nextInt();
        long S = fs.nextInt();
        long P = fs.nextInt();
        StringBuilder T = new StringBuilder(fs.next());

        long[] dp = new long[N+1];

        char goo = 'r';
        char choki = 's';
        char par = 'p';

        for (int i = 1; i <= N; i++) {
            char c = T.charAt(i-1);

            if (i - 1 - K >= 0) {
                char prev = T.charAt(i - 1 - K);
                long score = c == choki? R : c == par ? S : c == goo ? P : 0;
                if (prev == c) {
                    score = 0;
                    T.setCharAt(i-1, 'X');
                }
                dp[i] = score + dp[i-1];
            } else {
                long score = c == choki? R : c == par ? S : c == goo ? P : 0;
                dp[i] = score + dp[i-1];
            }
        }

        out.println(dp[N]);
        out.flush();
    }

    static void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    // Execute func for all patterns can be represented with n bit.
    // O(n2^n)
    static void forAllBit(int n, Consumer<List<Boolean>> func) {
        for (int i = 0; i < (1 << n); i++) {
            List<Boolean> arg = new ArrayList();
            for (int j = 0; j < n; j++) {
                boolean isJthBitSet = ((i >> j) & 1) == 1;
                arg.add(isJthBitSet);
            }
            func.accept(arg);
        }
    }

    static class Pair<F extends Comparable<F>, S extends Comparable<S>> implements Comparable<Pair<F, S>> {
        F f;
        S s;

        Pair() {
        }

        Pair(F f, S s) {
            this.f = f;
            this.s = s;
        }

        Pair(Pair<F, S> p) {
            f = p.f;
            s = p.s;
        }

        @Override
        public int compareTo(Pair<F, S> p) {
            if (f.compareTo(p.f) != 0) {
                return f.compareTo(p.f);
            }
            return s.compareTo(p.s);
        }

        @Override
        public int hashCode() {
            return f.hashCode() ^ s.hashCode();
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || this.f == null || this.s == null) {
                return false;
            }
            if (this.getClass() != o.getClass()) {
                return false;
            }
            Pair p = (Pair) o;
            return this.f.equals(p.f) && this.s.equals(p.s);
        }

        @Override
        public String toString() {
            return "{" + f.toString() + ", " + s.toString() + "}";
        }
    }

    public static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    public static boolean isPrime(int n) {
        for (int i = 2; i <= ((int) Math.sqrt(n)); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return n != 1;
    }

    public static void reverse(int[] array) {
        for (int i = 0; i < array.length / 2; i++) {
            int temp = array[i];
            array[i] = array[array.length - i - 1];
            array[array.length - i - 1] = temp;
        }
    }

    public static void reverse(long[] array) {
        for (int i = 0; i < array.length / 2; i++) {
            long temp = array[i];
            array[i] = array[array.length - i - 1];
            array[array.length - i - 1] = temp;
        }
    }

    public static boolean isPrime(long n) {
        for (long i = 2; i <= ((long) Math.sqrt(n)); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return n != 1;
    }

    public static Map<Long, Long> primeFactor(long n) {
        Map<Long, Long> map = new HashMap<>();
        long num = n;
        for (long i = 2; i <= ((long) Math.sqrt(n)); i++) {
            while (num % i == 0L) {
                map.put(i, map.getOrDefault(i, 0L) + 1L);
                num /= i;
            }
        }
        if (num != 1) {
            map.put(num, 1L);
        }

        return map;
    }

    public static List<Long> divisors(long n) {
        List<Long> list = new ArrayList<>();
        for (long i = 1; i <= ((long) Math.sqrt(n)); i++) {
            if (n % i == 0) {
                list.add(i);
                if (i != n / i) {
                    list.add(n / i);
                }
            }
        }

        return list;
    }

    public static int lcm(int a, int b) {
        return a / gcd(a, b) * b;
    }

    public static int nCk(int n, int k, int M) {
        long ret = 1;
        int min = Math.min(k, n - k);
        for (int i = 1; i <= min; i++) {
            ret = (ret * pow(i, M - 2, M)) % M;
        }
        for (int i = n - min + 1; i <= n; i++) {
            ret = (ret * i) % M;
        }
        return (int) ret;
    }

    //return a^b mod M O(logB)
    public static long pow(long a, long b, int M) {
        long ret = 1;
        long tmp = a;
        while (b > 0) {
            if ((b & 1) == 1) ret = (ret * tmp) % M;
            tmp = (tmp * tmp) % M;
            b = b >> 1;
        }
        return ret;
    }


    public static int find(int[] tree, int idx) {
        if (tree[idx] == idx) return idx;
        else return tree[idx] = find(tree, tree[idx]);
    }

    public static void union(int[] tree, int idx1, int idx2) {
        int root1 = find(tree, idx1);
        int root2 = find(tree, idx2);
        tree[root2] = root1;
    }

    public static int lowerBound(int[] list, int target) {
        int ok = list.length;
        int ng = -1;
        while (Math.abs(ok - ng) > 1) {
            int mid = (ok + ng) / 2;
            if (list[mid] >= target) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        return ok;
    }

    public static int lowerBound(long[] list, long target) {
        int ok = list.length;
        int ng = -1;
        while (Math.abs(ok - ng) > 1) {
            int mid = (ok + ng) / 2;
            if (list[mid] >= target) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        return ok;
    }

    public static <T extends Comparable<? super T>> int lowerBound(List<T> list, T target) {
        return ~Collections.binarySearch(list, target, (x, y) -> (x.compareTo(y) >= 0) ? 1 : -1);
    }

    public static <T extends Comparable<? super T>> int upperBound(List<T> list, T target) {
        return ~Collections.binarySearch(list, target, (x, y) -> (x.compareTo(y) > 0) ? 1 : -1);
    }
}


class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte()) return buffer[ptr++];
        else return -1;
    }

    private boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }
}

class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    /**
     * 1-indexed なBinary Indexed Treeを構築する
     *
     * @param n   容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    BIT(int n, BiFunction<T, T, T> bif, Supplier<T> sup) {
        this.n = n;
        bit = new ArrayList<>(n + 1);
        for (int i = 0; i < n + 1; ++i) {
            bit.add(sup.get());
        }
        this.bif = bif;
    }

    /**
     * iの位置の値をvで更新する
     *
     * @param i index
     * @param v 新しい値
     */
    void set(int i, T v) {
        for (int x = i; x <= n; x += x & -x) {
            bit.set(x, bif.apply(bit.get(x), v));
        }
    }

    /**
     * クエリー
     *
     * @param defaultValue 初期値
     * @param i            index
     * @return [1, i]までfを適用した結果
     */
    T reduce(T defaultValue, int i) {
        T ret = defaultValue;
        for (int x = i; x > 0; x -= x & -x) {
            ret = bif.apply(ret, bit.get(x));
        }
        return ret;
    }
}

class SegmentTree<T> {
    int n;
    ArrayList<T> dat;
    BiFunction<T, T, T> bif;
    Supplier<T> sup;

    /**
     * 0-indexed なSegment Treeを構築する
     *
     * @param n_  要求容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    SegmentTree(int n_, BiFunction<T, T, T> bif, Supplier<T> sup) {
        n = 1;
        while (n < n_) n *= 2;

        dat = new ArrayList<>(2 * n - 1);
        for (int i = 0; i < 2 * n - 1; ++i) {
            dat.add(sup.get());
        }
        this.bif = bif;
        this.sup = sup;
    }

    /**
     * kの位置の値をvで更新する
     *
     * @param k index
     * @param v 新しい値
     */
    void set(int k, T v) {
        k += n - 1;
        dat.set(k, v);
        while (k > 0) {
            k = (k - 1) / 2;
            dat.set(k, bif.apply(dat.get(k * 2 + 1), dat.get(k * 2 + 2)));
        }
    }

    /**
     * クエリー
     *
     * @param l はじめ
     * @param r おわり
     * @return [l, r)での演算bifを適用した結果を返す
     */
    T reduce(int l, int r) {
        return _reduce(l, r, 0, 0, n);
    }

    T _reduce(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return sup.get();
        if (a <= l && r <= b) return dat.get(k);
        T vl = _reduce(a, b, k * 2 + 1, l, (l + r) / 2);
        T vr = _reduce(a, b, k * 2 + 2, (l + r) / 2, r);
        return bif.apply(vl, vr);
    }
}

class UnionFind {
    int[] par;

    UnionFind(int n) {
        par = new int[n];
        for (int i = 0; i < n; ++i) {
            par[i] = i;
        }
    }

    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    boolean same(int x, int y) {
        return find(x) == find(y);
    }

    void union(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        par[x] = y;
    }
}