/*
https://zhuanlan.zhihu.com/p/588646549
*/
#include <bits/stdc++.h>
#define int long long

using namespace std;

const int inf = 1E18;

inline int read () {
    int w = 1, s = 0; char ch = getchar ();
    for (; ! isdigit (ch); ch = getchar ()) if (ch == '-') w = -1;
    for (; isdigit (ch); ch = getchar ()) s = (s << 1) + (s << 3) + (ch ^ 48);
    return s * w;
}

void solve () {
	int n = read (), k = read ();
    
    vector <int> a (n + 1), b (n + 1);
    for (int i = 1; i <= n; i ++) {
        a[i] = read ();
    }
    for (int i = 1; i <= n; i ++) {
        b[i] = read ();
    }

    sort (a.begin () + 1, a.end (), greater ());
    sort (b.begin () + 1, b.end (), greater ());

    vector dp (2 * n + 1, vector (k + 1, vector <int> (k + 1)));
    vector vis (2 * n + 1, vector (k + 1, vector <bool> (k + 1)));

    auto dfs = [&] (auto &&self, int x, int c1, int c2) {
        if (x == n * 2 + 1) {
            return 0LL;
        }
        if (vis[x][c1][c2]) {
            return dp[x][c1][c2];
        }
        vis[x][c1][c2] = true;
        int &ans = dp[x][c1][c2];
        int r1 = x / 2, r2 = (x - 1) / 2;
        int b1 = r2 - c2, b2 = r1 - c1;
        int p1 = c1 + b1 + 1, p2 = c2 + b2 + 1;

        if (x & 1) {
            ans = max (
                p2 <= n ? -self (self, x + 1, c1, c2) : -inf,
                p1 <= n && c1 < k ? a[p1] - self (self, x + 1, c1 + 1, c2) : -inf
            );
        } else {
            ans = max (
                p1 <= n ? -self (self, x + 1, c1, c2) : -inf,
                p2 <= n && c2 < k ? b[p2] - self (self, x + 1, c1, c2 + 1) : -inf
            );
        }
        return ans;
    };
    printf ("%lld\n", dfs (dfs, 1, 0, 0));
} 

signed main () {
	// ios::sync_with_stdio (false);
    // cin.tie (nullptr);
    
	int T = 1; 
	// cin >> T;
	// T = read ();

	while (T --) {
		solve ();
	}
	
	return 0;
}
/*
dp[i][j][k] i轮a选j个，b选k个,接下来的最大分差
然后要求等价于己方分-对方分最大，所以自己要选自己最大，ban对方最大
进行i轮之前，a进行了t1 = (i) / 2轮，b进行了t2 = (i - 1) / 2轮
对于a，由于b选了k个，则ban了a (t2 - k)个，所以对于将要选的第j个是a的第(j + (t2 - k) + 1)大
对于b同理
最后注意选不能大于k，ban不能大于n
然后进行记忆化搜索
*/