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())
MOD = int(1e9+7)
dp = [1]
for i in range(1,min(k+1,n+1)):
    dp.append(2**(i-1)%MOD)
for i in range(k+1,n+1):
    s = 0
    for j in range(i-1,i-k-1,-1):
        s = (s%MOD + dp[j]%MOD)%MOD
    dp.append(s)

#print(dp)
first_val = dp[-1]

k = d-1
dp = [1]
for i in range(1,min(k+1,n+1)):
    dp.append(2**(i-1)%MOD)
for i in range(k+1,n+1):
    s = 0
    for j in range(i-1,i-k-1,-1):
        s = (s%MOD + dp[j]%MOD)%MOD
    dp.append(s)
#print(dp)
second_val = dp[-1]

print((first_val%MOD - second_val%MOD)%MOD)
