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++14 (Clang 3.8.0) (judge verdict: Accepted)

/**
 * Libraries
 */
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
#include <iomanip>
#include <set>
#include <unordered_map>

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

using namespace std;

/**
 * macros
 */
#define rep(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define repx(i,s,e,d) for(int i=s,i##_end=(e);i<i##_end;i+=d)
#define repxr(i,s,e,d) for(int i=s,i##_end=(e);i>i##_end;i+=d)
#define rept(n) for(int ___i___=0,i##_len=(n);___i___<i##_len;++___i___)
#define repi(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define ZERO(a) memset(a,0,sizeof(a))
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(v.begin(),v.end()),v.end());
#define CEIL(x, y) (((x) + (y) - 1) / (y))

/**
 * standard input helpers
 */
// vector
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec) is >> x;
    return is;
}
// pair
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}
// vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}
// map
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> &map_var) {
    os << "{";
    repi(itr, map_var) {
        os << *itr;
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set
template <typename T>
ostream &operator<<(ostream &os, set<T> &set_var) {
    os << "{";
    repi(itr, set_var) {
        os << *itr;
        itr++;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
/**
 * debug tools
 */
#define DUMPOUT cout
void dump_func() {
    DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) > 0) {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}
#ifdef LOCAL_
#define dump(...)                                                              \
    DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                            \
            << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]"        \
            << endl                                                            \
            << "    ",                                                         \
        dump_func(__VA_ARGS__)
#else
#define dump(...)
#endif

/**
 * helpers
 */
typedef long long int lli;
typedef pair<int, int> ii;
typedef priority_queue<int, vector<int>, greater<int> > heapq;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
struct aaa{
    aaa(){
        cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(20);
    };
}aaaaaaa;
// #define int long long int
// struct S{
//     bool operator<(const S& x) const {
//         return ;
//     }
// };
signed main() {
    int N, K; cin >> N >> K;
    int R, S, P; cin >> R >> S >> P;
    string T; cin >> T;

    int dp[100001][3]; // r, p, s
    rep(i, 100001) {
        rep(j, 3) {
            dp[i][j] = 0;
        }
    }

    lli ans = 0;
    for (int k = 0; k < K; k++) {
        lli score[3]; rep(i, 3) score[i] = 0;
        for (int i = k; i < N; i += K) {
            if (i < K) {
                if (T[i] == 'r') {
                    score[1] = P;
                } else if (T[i] == 'p') {
                    score[2] = S;
                } else {
                    score[0] = R;
                }
            } else {
                int rr = score[0], pp = score[1], ss = score[2];
                score[0] = max(pp, ss);
                score[1] = max(ss, rr);
                score[2] = max(pp, rr);
                if (T[i] == 'r') {
                    score[1] += P;
                } else if (T[i] == 'p') {
                    score[2] += S;
                } else {
                    score[0] += R;
                }
            }
            dump(i, score[0], score[1], score[2]);
        }
        lli maxScore = 0;
        rep (i, 3) {
            maxScore = max(score[i], maxScore);
        }
        ans += maxScore;
    }
    cout << ans << endl;
    return 0;
}