Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree.

A *k*-tree is an infinite rooted tree where:
 -  each vertex has exactly *k* children; -  each edge has some weight; -  if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1, 2, 3, ..., *k*. 
The picture below shows a part of a 3-tree.

 

 <img class="tex-graphics" src="https://espresso.codeforces.com/61ac5ac17c4d2f220968eff8fb59bb3c8dc77edf.png" style="max-width: 100.0%;max-height: 100.0%;"/>

 

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (10<sup class="upper-index">9</sup> + 7).
A single line contains three space-separated integers: *n*, *k* and *d* (1 ≤ *n*, *k* ≤ 100; 1 ≤ *d* ≤ *k*).
inp=input().split()
n=int(inp[0])
k=int(inp[1])
d=int(inp[2])
if (n==1 and d<=n):
    print(1)
elif (n==1 and d>n):
    print(0)
else:
    dp=[0 for i in range(n+1)]
    dp2=[0 for i in range(n+1)]
    dp[1]=1
    dp2[1]=1
    s1,s2,s3,s4,c,c2,k2=1,0,1,0,0,0,d-1
    for i in range(2,n+1):
        if i>k:
            dp[i]=s1-s2
            c=c+1
            s2=s2+dp[c]
        else:
            dp[i]=1+s1
        if (k2>0 and i>k2):
            dp2[i]=s3-s4
            c2=c2+1
            s4=s4+dp2[c2]
        elif k2>0:
            dp2[i]=1+s3
        s1=s1+dp[i]
        s3=s3+dp2[i]
    ans=(dp[n]-dp2[n])%1000000007
    print(ans)