// Original C++ solution (slower):
#include<bits/stdc++.h>

#define IL inline

#define RG register

#define LL long long

#define Upd(a) (a)>=mod?(a-=mod):1

using namespace std;

template <class T>

IL char gc () {

	static char buf[1<<6], *p1=buf, *p2=buf;

	return (p1==p2) && (p2=(p1=buf)+fread(buf,1,1<<6,stdin), p1==p2) ?EOF:*p1++;

}

template <typename T>

IL void read (T&data) {

	data=0;

	RG char ch=0;

	while (ch<'0' || ch>'9') ch=getchar();

	while (ch<='9' && ch>='0') data=(data<<1)+(data<<3)+(ch&15), ch=getchar();

}

const int mod = 1e9+7;

const int _ =2011;

int n, k;

int dp[2][_], jc[_*_], jcn[_*_];

IL int poww (RG int a, RG int b) {

	RG int ret=1, Base=a;

	for (;b;b>>=1, Base=1LL*Base*Base%mod) if (b&1) ret=1LL*ret*Base%mod;

	return ret;

}

IL int C (RG int a, RG int b) {

	if (a<b) return 0;

	return 1LL*jc[a]*jcn[b]%mod*jcn[a-b]%mod;

}

int main (/*int aa, int bb*/) {

	//freopen("1.in","r",stdin);

	//freopen("1.out","w",stdout);

	//n=aa, k=bb;

	read(n), read(k);

	if (k<=1) {puts("1"); return 0;}

	jc[0]=jcn[0]=1;

	RG int i, j, pre=1, cur=0;

	for (i=1; i<=n*k; ++i) jc[i]=1LL*jc[i-1]*i%mod, jcn[i]=poww(jc[i], mod-2);

	dp[0][0]=1;

	for (i=1; i<=n*2; ++i) {

		swap(cur, pre);

		for (j=0; j<min(i,n+1); ++j) dp[cur][j]=0;

		for (j=1; j<=min(i,n); ++j) {

			if ((i-j)%2) continue;

			if (n*2-i+1<j) break;

			RG int a=(i+j-2)>>1, b=(i-j)>>1;

			dp[cur][j]=1LL*dp[pre][j-1]*C(a*(k-1)+b+k-1-1,k-1-1)%mod;

			dp[cur][j]+=dp[pre][j+1];

			Upd(dp[cur][j]);

		}

		if (cur%2==0) dp[cur][0]=dp[pre][1];

	}

	//return 1LL*dp[cur][0]*jc[n]%mod;

	printf("%lld",1LL*dp[cur][0]*jc[n]%mod);

}


// Optimized solution to the same problem (faster):
#include <cstdio>



const int mod = 1e9 + 7;

const int MAXN = 2005, MAXS = MAXN * MAXN;



template<typename _T>

void read( _T &x )

{

	x = 0;char s = getchar();int f = 1;

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

	while( s >= '0' && s <= '9' ){x = ( x << 3 ) + ( x << 1 ) + ( s - '0' ), s = getchar();}

	x *= f;

}



template<typename _T>

void write( _T x )

{

	if( x < 0 ){ putchar( '-' ); x = ( ~ x ) + 1; }

	if( 9 < x ){ write( x / 10 ); }

	putchar( x % 10 + '0' );

}



int f[MAXN][MAXN];

int fac[MAXS], ifac[MAXS];

int N, K;



int qkpow( int, int );

int inv( const int a ) { return qkpow( a, mod - 2 ); }

void sub( int &x, const int v ) { x = ( x < v ? x - v + mod : x - v ); }

void add( int &x, const int v ) { x = ( x + v >= mod ? x + v - mod : x + v ); }



int qkpow( int base, int indx )

{

	int ret = 1;

	while( indx )

	{

		if( indx & 1 ) ret = 1ll * ret * base % mod;

		base = 1ll * base * base % mod, indx >>= 1;

	}

	return ret;

}



void init( const int siz )

{

	fac[0] = ifac[0] = 1;

	for( int i = 1 ; i <= siz ; i ++ ) fac[i] = 1ll * fac[i - 1] * i % mod;

	ifac[siz] = inv( fac[siz] );

	for( int i = siz - 1 ; ~ i ; i -- ) ifac[i] = 1ll * ifac[i + 1] * ( i + 1 ) % mod;

}



int C( const int n, const int m ) { return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod; }



int main()

{

	read( N ), read( K );

	if( K == 1 ) { puts( "1" ); return 0; }

	init( N * K );

	for( int i = 0 ; i <= N ; i ++ ) f[i][0] = 1;

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

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

		{

			f[i][j] = f[i - 1][j];

			int t = N * K - i - ( K - 1 ) * ( j - 1 ) - 1;

			add( f[i][j], 1ll * f[i][j - 1] * ( N - j + 1 ) % mod * C( t, K - 2 ) % mod );

		}

	write( f[N][N] ), putchar( '\n' );

	return 0;

}

// Fastest known solution for this problem: