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=map(int,input().split())
dp_array=[0 for i in range(100)]
# dp_array[k] tells the number of ways to reach k+1  
for i in range(k):
    dp_array[i]+=1
for j in range(n):
    for i in range(1,min(j+1,k+1)):
        dp_array[j]+=dp_array[j-i]
atleast_d=[0 for i in range(100)]
for i in range(d-1,k):
    atleast_d[i]+=1
for j in range(n):
    for i in range(1,min(j+1,d)):
        atleast_d[j]+=atleast_d[j-i]
    for i in range(d,min(j+1,k+1)):
        atleast_d[j]+=dp_array[j-i]
print(atleast_d[n-1])