bzoj3622: 已经没有什么好害怕的了

/*
首先解方程得到具体有多少个是大于的情况 
然后dp求出f[i]是至少有i个大于的情况
最后容斥一下就好了
 




*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define ll long long 
#define M 2020
#define mmp make_pair
using namespace std;
int read()
{
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
const int mod = 1000000009;

void add(int &x, int y)
{
	x += y;
	x -= x >= mod ? mod : 0;
	x += x < 0 ? mod : 0;
}

int mul(int x, int y)
{
	return 1ll * x * y % mod;
}


int f[M][M], a[M], b[M], c[M][M], fac[M], n, k; 
int main()
{
	n = read(), k = read();
	if((n - k) & 1) return 0 * puts("0");
	for(int i = 1; i <= n; i++) a[i] = read();
	for(int i = 1; i <= n; i++) b[i] = read();
	sort(a + 1, a + n + 1);
	sort(b + 1, b + n + 1);
	c[0][0] = 1, fac[0] = 1;
	for(int i = 1; i <= n; i++)
	{
		fac[i] = mul(fac[i - 1], i);
		c[i][0] = 1;
		for(int j = 1; j <= i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
	}
	f[0][0] = 1;
	for(int i = 1; i <= n; i++)
	{
		int tmp;
		for(tmp = 1; tmp <= n && b[tmp] < a[i]; tmp++);
		tmp--;
		for(int j = 1; j <= i; j++) f[i][j] = f[i - 1][j], add(f[i][j], mul(f[i - 1][j - 1], max(tmp - j + 1, 0)));
		f[i][0] = f[i - 1][0];
	}	
	int ans = 0;
	k = (n + k) >> 1;
	for(int i = k; i <= n; i++)
	{
		ll tmp = mul(f[n][i], mul(fac[n - i], c[i][k]));
		if((i - k) & 1) add(ans, -tmp);
		else add(ans, tmp);
	}
	cout << ans << "\n";
	return 0;
}