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*).

M= pow(10,9)+7

memo = {}
def clc(summ,flag):
    if summ == n:
        return flag == True
    if summ>n:return 0
    if (summ,flag) in memo:return memo[summ,flag]
    ans = 0
    for i in range(1,k+1):
        if(i>=d):ans +=clc(summ+i,True)%M
        else:ans+=clc(summ+i,flag)%M
    memo[summ,flag] = ans
    return ans

n,k,d = map(int,input().split())
ans = clc(0,False)
print(ans)
