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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
    int N, K;
    int R, S, P;
    char T[100001];
    char tmp;//K個前に出した手を保存する
    int tmp_can;//K個前の手を出すことが出来たかどうか(出せた場合→１、出せなかった→0)
    int i, j, r_count=0, s_count=0, p_count=0;//筐体が出した手をカウント
    long long int false_point=0, true_point=0, sub=0;//Kの被りを数える前と後、それから被り分のマイナスポイント

    scanf("%d %d",&N, &K);
    scanf("%d %d %d",&R, &S, &P);
    scanf("%s",T);

    //全部勝利した場合をカウント
    for(i=0; i<N; i++){
        if(T[i] == 'r'){
            r_count++;
        } else if(T[i] == 's'){
            s_count++;
        } else {
            p_count++;
        }
    }
    //K回の制限が無い場合のポイントが求まる
    false_point = r_count*P + s_count*R + p_count*S;


    r_count=0;
    s_count=0;
    p_count=0;

    //その後、K回目の被りを考えていく

    //i+j*kを考えていく方が良いのかも
    //tmp;  K個前に出した手を保存する
    //tmp_can;  K個前の手を出すことが出来たかどうか(出せた場合→１、出せなかった→0)

    for(i=0; i<K; i++){
        tmp_can=0;
        tmp = 'a';
        for(j=i; j<N; j=j+K){
            //printf("tmp=%c, T[%d]=%c\n",tmp, j, T[j]);
            if((j>K) && (tmp == T[j])){//K個前と被っている場合
                if(tmp_can == 1){//K個前でそれを出した場合（つまり、余剰分としてカウントしていく）
                    tmp_can =0;
                    if(T[i] == 'r'){
                        r_count++;
                    } else if(T[i] == 's'){
                        s_count++;
                    } else {
                        p_count++;
                    }
                } else {//K個前では出していなかった。しかし、K個後には出せない。
                    tmp_can = 1;
                }
            } else {
                tmp_can = 1;
            }
            tmp = T[j];
        }
    }


    sub = r_count*P + s_count*R + p_count*S;

    true_point = false_point - sub;

    printf("%lld\n",true_point);

    return 0;
}