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 permutePaths(n, k, d, seen, exclusion):
    for i in range(1, k + 1):
        if i < d and i < len(exclusion):
            exclusion[i] = 1
        if i < len(seen):
            seen[i] = 1

    for i in range(2, n + 1):
        for j in range(1, k + 1):
            if (i - j >= 0):
                seen[i] += seen[i - j]
            if (i - j >= 0 and j < d):
                exclusion[i] += exclusion[i - j]


def main():
    string_this = input()
    string_this = string_this.split(" ")
    n = int(string_this[0])
    k = int(string_this[1])
    d = int(string_this[2])
    seen = []
    exclusion = []
    for i in range(0, n + 2):
        seen += [0]
        exclusion += [0]
    permutePaths(n, k, d, seen, exclusion)
    print(seen[n] - exclusion[n] % 1000000007)


main()

	 	 				     				     	    		