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*).
#author: sushmanth

from sys import stdin
input = stdin.readline

inp = lambda : list(map(int,input().split()))

mod = (10 ** 9) + 7
def answer():

    dp = [[0 , 0] for i in range(n + 1)]
    dp[0][0] = 1

    for i in range(1 , n + 1):

        for j in range(k , 0 , -1):
            if(i - j < 0):continue
            if(j < d):
                dp[i][0] += dp[i - j][0]
                dp[i][0] %= mod
            else:
                dp[i][1] += dp[i - j][0]
                dp[i][1] %= mod

            dp[i][1] += dp[i - j][1]
            dp[i][1] %= mod


    return dp[n][1]

for T in range(1):

    n , k , d = inp()

    print(answer())
            

    
   

    
