Problem:
Problem Statement

Snuke loves colorful balls. He has a total of N×K balls, K in each of his favorite N colors. The colors are numbered 1 through N.

He will arrange all of the balls in a row from left to right, in arbitrary order. Then, for each of the N colors, he will paint the leftmost ball of that color into color 0, a color different from any of the N original colors.

After painting, how many sequences of the colors of the balls are possible? Find this number modulo 10^9+7.

Constraints

1≤N,K≤2,000

Input

The input is given from Standard Input in the following format:

N K

Output

Print the number of the possible sequences of the colors of the balls after painting, modulo 10^9+7.

Sample Input 1

2 2

Sample Output 1

4

The following 4 sequences are possible:

(0,1,0,2)

(0,0,1,2)

(0,2,0,1)

(0,0,2,1)

Sample Input 2

3 1

Sample Output 2

1

The following 1 sequence is possible:

(0,0,0)

Sample Input 3

2 3

Sample Output 3

14

Sample Input 4

2000 2000

Sample Output 4

546381702

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include"bits/stdc++.h"


#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
#define iout(x) printf("%d\n",x)
#define lout(x) printf("%lld\n",x)
#define REP(x,l,u) for(ll x = (l);x<=(u);x++)
#define RREP(x,l,u) for(ll x = (l);x>=(u);x--)
#define mst(x,a) memset(x,a,sizeof(x))
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define MP make_pair
#define se second
#define fi first
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
#define sz(x) ((int)x.size())
#define cl(x) x.clear()

typedef  long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
using namespace std;

const int maxn = 2010;
const int mod = 1e9+7;
const double eps = 1e-6;
const double PI = acos(-1);

template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}

template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}
template<typename A,typename B> inline A fexp(A x,B p){A ans=1;for(;p;p>>=1,x=1LL*x*x%mod)if(p&1)ans=1LL*ans*x%mod;return ans;}
template<typename A,typename B> inline A fexp(A x,B p,A mo){A ans=1;for(;p;p>>=1,x=1LL*x*x%mo)if(p&1)ans=1LL*ans*x%mo;return ans;}

int n,k;

int f[maxn][maxn];

int fac[maxn*maxn],inv[maxn*maxn];

int Cnk(int n,int k){
	return 1LL*fac[n]*inv[k]%mod*inv[n-k]%mod;
}

void Work(){
	if(k==1)return void(puts("1"));
	f[n+1][n+1]=1;
	RREP(i,n+1,1)
		RREP(j,n,1)if(i>=j){
			if(i<=n)f[i][j]=f[i+1][j];
			if(i>j)f[i][j]=(f[i][j]+1LL*f[i][j+1]*Cnk(n-i+(n-j+1)*(k-1),k-2))%mod;
			//cout<<i<<' '<<j<<' '<<f[i][j]<<endl;
		}
	REP(i,1,n)f[1][1]=1LL*f[1][1]*i%mod;
	iout(f[1][1]);
}

void Init(){
	read(n,k);
	fac[0]=1;REP(i,1,n*k)fac[i]=1LL*fac[i-1]*i%mod;
	REP(i,0,n*k)inv[i]=fexp(fac[i],mod-2);
}

int main(){
	Init();
	Work();
	return 0;
}