题意：和上题基本一样，求至少k对a>b的方案数。不取模！！！

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做k+1遍容斥就行了

高精度超强！！！几乎把所有的都用上了

然后，注意有负数，所以容斥的时候正负分别保存然后再一减就行了

~~这是我省选前最后一次写高精度了~~

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long ll;

const int N=205, B=1e4;

inline int read(){

	char c=getchar();int x=0,f=1;

	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}

	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}

	return x*f;

}

struct Big{

	int a[120], n;

	int& operator [](int x) {return a[x];}

	Big():n(1) {memset(a, 0, sizeof(a));}

	void ini(int x) {a[1]=x; n=1;}

}f[N][N], C[N][N], fac[N];

Big operator *(Big a, int b) {

	int g=0;

	for(int i=1; i<=a.n; i++)

		g += a[i]*b, a[i] = g%B, g/=B;

	if(g) a[++a.n] = g;

	return a;

}

Big operator *(Big a, Big b) {

	Big c;

	for(int i=1; i<=a.n; i++) {

		int g=0;

		for(int j=1; j<=b.n; j++)

			g += c[i+j-1]+a[i]*b[j], c[i+j-1] = g%B, g/=B;

		c[i+b.n] = g;

	}

	c.n = a.n + b.n;

	while(c.n>1 && c[c.n]==0) c.n--;

	return c;

}

Big operator +(Big a, Big b) {

	int g=0, n=max(a.n, b.n);

	for(int i=1; i<=n; i++) {

		g += i<=a.n ? a[i] : 0;

		g += i<=b.n ? b[i] : 0;

		a[i] = g%B, g/=B;

	}

	a.n = n;

	if(g) a[++a.n] = g;

	return a;

}

Big operator -(Big a, Big b) {

	for(int i=1; i<=b.n; i++) {

		if(a[i]<b[i]) a[i]+=B, a[i+1]--;

		a[i] -= b[i];

	}

	int p=b.n+1;

	while(a[p]<0) a[p]+=B, a[++p]--;

	while(a.n>1 && a[a.n]==0) a.n--;

	return a;

}

void Print(Big &a) {

	printf("%d", a[a.n]);

	for(int i=a.n-1; i>=1; i--) printf("%04d", a[i]);

}

int n, k, a[N], b[N];

int g[N];

void dp() {

	fac[0].ini(1);

	for(int i=1; i<=n; i++) fac[i] = fac[i-1]*i;

	C[0][0].ini(1);

	for(int i=1; i<=n; i++) {

		C[i][0].ini(1);

		for(int j=1; j<=n; j++) C[i][j] = C[i-1][j] + C[i-1][j-1];

	}

	int now=0;

	for(int i=1; i<=n; i++) {

		while(now<n && a[i]>b[now+1]) now++;

		g[i] = now;

	}

	for(int i=0; i<=n; i++) f[i][0].ini(1);

	for(int i=1; i<=n; i++)

		for(int j=1; j<=g[i]; j++) f[i][j] = f[i-1][j] + f[i-1][j-1]*(g[i]-j+1);// Print(f[i][j]), puts("");

}

int main() {

	freopen("in","r",stdin);

	n=read(); k=read();

	for(int i=1; i<=n; i++) b[i]=read();

	for(int i=1; i<=n; i++) a[i]=read();

	sort(a+1, a+1+n); sort(b+1, b+1+n);

	dp();

	Big pos, neg, ans;

	int kkk=k;

	for(int k=0; k<=kkk; k++) { //printf("kkkkkk %d\n",k);

		for(int i=k; i<=n; i++){ //printf("i %d\n",i);

			Big t = fac[n-i] * f[n][i] * C[i][k];  //printf("t ");Print(t); puts("");

			if((i-k)&1) neg = neg + t;

			else pos = pos + t;

		}

	}

	ans = pos - neg;

			//printf("hi %d  ",i),Print(ans), puts("");

	Print(ans);

}