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 - target sum
# k - number of child
# d - path must contain d
import sys

sys.setrecursionlimit(1000000007)

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



memo = {(0, True): 1, (0, False): 0}


def counter(left_sum, contains_d):
    if (left_sum, contains_d) in memo:
        return memo[(left_sum, contains_d)]


    path_count = 0
    for i in range(1, k + 1):
        if i <= left_sum:            
            path_count += counter(left_sum - i, contains_d or i >= d)        
    
    memo[(left_sum, contains_d)] = path_count
    
    return path_count

path_count = counter(n, False)

print(path_count %  1000000007 )