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*).
MOD = 1e9+7

def add(a,b):
    a += b
    if a>=MOD:
        a-=MOD
    return a
   
   
def solve():
    inp = input().split()
    inp = [int(i) for i in inp]
    n,k,d = inp
    dp = [[0,0] for i in range(10)]
    dp[0][0] = 1
    for i in range(1,n+1):
        for j in range(1,min(k+1, i+1)):
            if j>=d:
                dp[i][1] = add(dp[i][1],dp[i-j][1])
                dp[i][1] = add(dp[i][1], dp[i-j][0])
            else:
                dp[i][1] = add(dp[i][1], dp[i-j][1])
                dp[i][0] = add(dp[i][0], dp[i-j][0])
    # print(dp)
    print(dp[n][1])
            
    
    
    
    
def main():
    # t = int(input())
    t = 1
    for _ in range(t):
        solve()



if __name__ == "__main__":
    main()
    
    
# elif ((i==n and i==d) or (i==n and d == 1) or (i==n and d != actualD)):
# elif((i==n) and ((i==d) or (d==1) or (d != actualD))):
