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*).
from functools import lru_cache
def numI(): return int(input())
def numII(): return list(input())
def numIII(): return list(map(int, input().split(' ')))
def strInp(): return input()
def strInpI(): return list(input())
@lru_cache
def solve():
    n, k, d = numIII()
    c = 0
    memo = {}
    def dfs(num, maxi):
        nonlocal c
        if num == n and maxi >= d:
            c+=1
            return 1
        # if num in memo:
        #     return memo[num]
        temp = 0
        for x in range(1, k+1):
            if x + num <= n:
                ret = dfs(x + num, max(maxi, x))
                temp+=ret
        memo[num] = temp
        return temp
    dfs(0, -1)
    print(c)
solve()