1 #include <cstdio>
 2 #include <cstring>
 3 #include <stack>
 4 #include <algorithm>
 5 
 6 #define ll long long
 7 int const N = 1005;
 8 ll const mod = 1000000007;
 9 
10 using namespace std;
11 
12 ll stl[N][N];
13 ll sumstl;
14 ll n,r,k,m;
15 ll dp[N][N];
16 ll sum[N][N];
17 ll ans;
18 
19 void ini1()
20 {
21     memset(stl,0,sizeof(stl));
22     ll i;
23     for(i=1;i<=1000;i++){
24         stl[i][i]=1;
25     }
26     stl[0][0]=1;
27     ll p,j;
28     for(p=1;p<=1000;p++){
29         for(j=1;j<=p;j++){
30             stl[p][j]= (j*stl[p-1][j]+stl[p-1][j-1])%mod;
31         }
32     }
33 
34     //for(p=1;p<=10;p++){
35    //     for(j=1;j<=p;j++) printf(" p=%I64d j=%I64d stl=%I64d\n",p,j,stl[p][j]);
36    // }
37 }
38 
39 void ini()
40 {
41     ll i;
42     sumstl=0;
43     for(i=1;i<=m;i++){
44         sumstl=(sumstl+stl[r][i])%mod;
45     }
46     memset(dp,0,sizeof(dp));
47     memset(sum,0,sizeof(sum));
48     ans=0;
49    // printf(" sumstl=%I64d\n",sumstl);
50 }
51 
52 void solve()
53 {
54     int i,j;
55     for(i=n;i>=1;i--){
56         dp[i][1]=1;
57         sum[i][1]=(sum[i+1][1]+dp[i][1])%mod;
58     }
59     for(j=2;j<=r;j++){
60         for(i=n-k;i>=1;i--){
61             dp[i][j]=sum[i+k][j-1];
62             sum[i][j]=(sum[i+1][j]+dp[i][j])%mod;
63         }
64     }
65     ans=(sum[1][r]*sumstl)%mod;
66 }
67 
68 void out()
69 {
70     printf("%I64d\n",ans);
71 }
72 
73 int main()
74 {
75     //freopen("data.in","r",stdin);
76     ini1();
77     while(scanf("%I64d%I64d%I64d%I64d",&n,&r,&k,&m)!=EOF)
78     {
79         ini();
80         solve();
81         out();
82     }
83 }