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 types import GeneratorType
from collections import defaultdict as dd
def bootstrap(f,queue=[]):
    def wrappedfunc(*args,**kwargs):
        if queue:return f(*args, **kwargs)
        else:
            to=f(*args, **kwargs)
            while True:
                if isinstance(to,GeneratorType):queue.append(to);to=next(to)
                else:
                    queue.pop()
                    if not queue:break
                    to=queue[-1].send(to)
            return to
    return wrappedfunc
n,k,d=map(int,input().split())
MOD=10**9+7
@bootstrap
def f(s,mx):
    if s>n:yield 0
    if s==n:
        if mx>=d:yield 1
        yield 0
    cur=0
    for i in range(1,k+1):
        x=f(s+i,max(mx,i))
        x=yield x
        cur=(cur+x)%MOD
    yield cur
x=f(0,0)
print(x)