#include <iostream>
#include <vector>

using namespace std;
using ll = long long;

int main () {
    // 色と場所を持って区間dpできそう

    int N, K; cin >> N >> K;
    vector<int> A(N), C(N);

    for (int i = 0; i < N; i++) cin >> A[i];
    for (int i = 0; i < N; i++) cin >> C[i];
    for (int i = 0; i < N; i++) C[i]--;

    vector<vector<vector<ll>>> dp(N, vector<vector<ll>>(N+1, vector<ll>(50, 0)));
    // dp[i][j][c] := iから右回りにj個の区間を全統合し、結果色cの石を作ったとき、価値の最大

    for (int i = 0; i < N; i++) dp[i][1][C[i]] = A[i];

    for (int i = 0; i < N; i++) {
        for (int j = 2; j <= N; j++) {
            for (int c = 0; c < 50; c++) {
                for (int k = 0; k <= K; k++) {
                    for (int mid = 1; mid < j; mid++) {
                        if (0 < dp[i][mid][c] && 0 < dp[(i+mid)%N][j-mid][min(49, c+k)]) {
                            dp[i][j][c] = max(dp[i][j][c], dp[i][mid][c] + dp[(i+mid)%N][j-mid][min(49, c+k)]);
                        }

                        if (0 < dp[i][mid][c] && 0 < dp[(i+mid)%N][j-mid][max(0, c-k)]) {
                            dp[i][j][c] = max(dp[i][j][c], dp[i][mid][c] + dp[(i+mid)%N][j-mid][max(0, c-k)]);
                        }

                        if (0 < dp[i][mid][min(49, c+k)] && 0 < dp[(i+mid)%N][j-mid][c]) {
                            dp[i][j][c] = max(dp[i][j][c], dp[i][mid][min(49, c+k)] + dp[(i+mid)%N][j-mid][c]);
                        }

                        if (0 < dp[i][mid][c] && 0 < dp[(i+mid)%N][j-mid][max(0, c-k)]) {
                            dp[i][j][c] = max(dp[i][j][c], dp[i][mid][max(0, c-k)]  + dp[(i+mid)%N][j-mid][c]);
                        }
                    }
                }
            }
        }
    }

    ll ans = 0;
    for (int i = 0; i < N; i++) for (int j = 1; j <= N; j++) for (int c = 0; c < 50; c++) ans = max(ans, dp[i][j][c]);

    cout << ans << "\n";
}