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*).
n, k, d = map(int, input().split())
dpNoRestriction = [0 for _ in range (n+1)]
dpNoRestriction[0] = 0
dpNoRestriction[1] = 1
for i in range (2,n+1) :
    if i<=k : dpNoRestriction[i] = 1
    for j in range (1,k+1) :
        if i-j>= 0 :
            dpNoRestriction[i] += dpNoRestriction[i-j]
# print(dpNoRestriction)

dpRestriction = [0 for _ in range (n+1)]
dpRestriction[d] = 1
for i in range (d+1, n+1) :
    if i<=k : dpRestriction[i] = 1
    for j in range (1, k+1) :
        if i-j >= 0 :
            if j>=d : dpRestriction[i] += dpNoRestriction[i-j]
            else : dpRestriction[i] += dpRestriction[i-j]
print(dpRestriction[-1])