"""
1<=n,d<=300
n<=k<=n*d
貰うdp
x段目に到達できるのは前回x-d~x-1段目にいるとき
累積和で高速化
"""
n,d,k=map(int,input().split())
dp=[0]*(n*d+1)
dp[0]=1
mod=10**9+7
for _ in range(n):
  ndp=[0]*(n*d+1)
  tmp=0
  for i in range(n*d+1):
    if 0<=i-d-1:
      tmp-=dp[i-d-1]
      tmp%=mod
    ndp[i]=tmp # sum(dp[i-d:i])
    tmp+=dp[i]
    tmp%=mod
  dp=ndp
print(dp[k])