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*).
from functools import lru_cache

@lru_cache(None)
def solve(n,k,d,f):
    # base case
    if(n == 0 and f):
        return 1
    if(n < 0):
        return 0
    if(n == 0 and not f):
        return 0
    ans = 0
    for i in range(1,k+1):
        if(f):
            ans += solve(n-i,k,d,f)
        else:
            if(i>=d):
                f = 1
            ans += solve(n-i,k,d,f)
    return ans

n,k,d = [int(ele) for ele in input().split()]
ans = solve(n,k,d,0)
print(ans)