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 solve_rec(n, k, d, has_d_edge):
    paths = 0
    for i in range(1,k+1):
        if n-i < 0:
            continue
        elif n-i == 0:
            if has_d_edge or i >= d:
                paths += 1
        else:
            if i >= d:
                has_d_edge = True
            paths += solve_rec(n-i, k, d, has_d_edge)
    return paths

n, k, d = map(int, input().split())
dp = {}

def call_solve():
    return solve(n, 0)

def solve(n, has_d_edge):
    if n < 0:
        return 0
    if n == 0:
        return has_d_edge
    if (n, has_d_edge) in dp.keys():
        return dp[(n, has_d_edge)]

    num_of_paths = 0
    for k_i in range(1,k+1):
        if k_i < d:
            num_of_paths += solve(n-k_i, has_d_edge)
        else: # k_i >= d
            num_of_paths += solve(n-k_i, 1)
    num_of_paths %= 1000000007
    dp[(n, has_d_edge)] = num_of_paths
    return num_of_paths

print(call_solve())

					  	    	 		    		  					 	