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*).
def rl():
    return list(map(int,input().split()))
def ri():
    return int(input())
def rm():
    return map(int,input().split())
def main():
    n,k,d=rm()
    if n<d: return 0
    if n==d: return 1
    a=[[0 for i in range(d+1)] for j in range(n+1)]
    for i in range(n+1):
        if i<d+1: a[i][i]=1; a[0][i]=1  
           
    for i in range(2,n+1):
        for j in range(1,k+1):
            if i-j>=0: a[i][1]+=a[i-j][1]
    for i in range(d+1,n+1):
        for j in range(1,d):
            if i-j>=0: a[i][d]+=a[i-j][d]
        for j in range(d,k+1):
            if i-j>=0: a[i][d]+=a[i-j][1]
    return a[n][d]
print(main())



