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 func(v, k, d, memo):
    if v in memo:
        return memo[v]
    else:
        memo[v] = [0, 0]
        nb = 0
        t = 0
        for i in range(1, k+1):
            if v-i > 0:
                ti, nbi = func(v-i, k, d, memo)
                if i >= d:
                    ti = nbi

            elif v-i == 0:
                if i >= d:
                    ti = 1
                    nbi = 1
                else:
                    ti = 0
                    nbi = 1

            else:
                break

            t += ti
            nb += nbi

        memo[v] = [t, nb]

        return memo[v]


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

memo = {}
a, b = func(n, k, d, memo)

print(a)
