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 ensurepip import bootstrap


n, k , d = map(int, input().split())
mod = 10**9 + 7

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

memo = {}
@bootstrap
def dp(cur, d_taken):
    if (cur, d_taken) in memo: return memo[cur, d_taken]
    if cur > n:
        return 0
    
    if cur == n and d_taken:
        return 1
    
    res = 0
    for choice in range(1, k+1):
        if choice >= d:
            res += (yield dp(cur + choice, True))%mod
        else:
            res += (yield dp(cur + choice, d_taken))%mod
    
    memo[cur, d_taken] = res%mod
    return memo[cur, d_taken]

print(dp(0, False))