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*).

L = input()
L = L.split()
S = [int(x) for x in L]

n = S[0] # нужный вес пути
k = S[1] # диапазон веса ребра
d = S[2] # хотя бы одно ребро этого веса

dd = d - 1

s = 'F(b-1)'
i = 2
while i <= k:
    s = s + ' + F(b-' + f'{i}' + ')'
    i += 1

print(s)

ss = 'FF(b-1)'
i = 2
while i < d:
    ss = ss + ' + FF(b-' + f'{i}' + ')' 
    i += 1

print(ss)


def F(b):
    if b == 0:
        return 1
    elif b < 0:
        return 0
    else:
        return eval(s)


def FF(b):
    if b == 0:
        return 1
    elif b < 0:
        return 0
    else:
        return eval(ss)

    
print(F(n) - FF(n))
