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 bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = stack.append(f(*args, **kwargs))
            while True:
                try:
                    to = stack.append(stack[-1].send(to))
                except StopIteration as e:
                    stack.pop()
                    to = e.value
                    if not stack:
                        break
            return to
 
    return wrappedfunc
 
@bootstrap
def go(curs,ddc):
    # print(curs,ddc)
    if (curs==n):
        if (ddc==1):
            return 1 
        return 0 
    if (curs>n):
        return 0 
    if (dp[ddc][curs]!=-1):
        return dp[ddc][curs] 

    c = 0 
    for i in range(1,k+1):
        if (i>=d):
            c+=(yield go(curs+i,ddc|1)) 
        else:
            c+=(yield go(curs+i,ddc)) 
    dp[ddc][curs]=c%(10**9+7) 
    return c%(10**9+7) 

    







n,k,d = map(int,input().split()) 
ans = 0 
dp = [ ] 
for i in range(2):
    x = [ ] 
    for j in range(n+1):
        x.append(-1) 
    dp.append(x) 

print(go(0,0)) 
