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*).
n,k,d = [int(e) for e in input().split() ]
memory = [[-1]*k]*101
count = 0 
if n < d:
    print(0)
else:
    D = [[1]*(n+1)]*2
    D[0][0] = 1
    for i in range(1,n+1): 
        for j in range(1, min(i+1 , d )):
            D[0][i] += D[0][i - j]
    for i in range(d):
        D[1][i]  = 0 
    D[1][d] = 1 
    for i in range(d+1 , n+1):
        for j in range(1,  min(k+1,n+1) ):
            if j < d : 
                D[1][i] += D[1][i - j]
            else: 
                D[1][i] += D[0][i-j] + D[1][i-j]
        # for j in range(d,min(k+1,n+1) ): 
        #     D[1][i] += 
print(D[1][n]%1000000007)
