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*).
n, k, d = [int(x) for x in input().split()]

f = [[0 for j in range(n + 1)] for i in range(n + 1)]
g = [[0 for j in range(n + 1)] for i in range(n + 1)]

#f[x][y] = the number of paths length x with weight y

f[0][0], g[0][0] = 1, 1

f_count, g_count = 0, 0 
for x in range(n):
    for y in range(x, n):
        if f[x][y] == 0:
            break

        for diff in range(1, k + 1):
            try:
                f[x + 1][y + diff] += f[x][y]
                if y + diff == n: f_count += (f[x][y])
            except:
                break

        for diff in range(1, d):
            try:
                g[x + 1][y + diff] += g[x][y]
                if y + diff == n: g_count += (g[x][y])
            except:
                break

print((f_count - g_count) % (10 ** 9 + 7))







    