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*).
import sys
from collections import deque, defaultdict, Counter
from functools import lru_cache

############ constants ############
MOD = 1000000007

############ Input Functions ############
input = sys.stdin.readline

def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

############ Output Functions ############
def oupNum(n):
    sys.stdout.write(str(n) + "\n")

def oupList(lst):
    sys.stdout.write(" ".join(map(str, lst)) + "\n")


############ Recursion Trick ############ 
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = stack.append(f(*args, **kwargs))
            while True:
                try:
                    to = stack.append(stack[-1].send(to))
                except StopIteration as e:
                    stack.pop()
                    to = e.value
                    if not stack:
                        break
            return to
    return wrappedfunc


def main():
    @bootstrap
    def dfs(rest, k, depth, must, hasMust):
      if rest < 0: return 0
      if rest == 0 and hasMust: return 1
      if rest == 0: return 0
      
      if (rest, depth, hasMust) in memo:
        return memo[(rest, depth, hasMust)]
  
      answer = 0
      for n in range(1, k+1):
        answer += yield dfs(rest-n, k, depth+1, must, hasMust or n >= must)
      memo[(rest, depth, hasMust)] = answer
      return answer

    total, k, must = inlt()
    memo = {}
    oupNum(dfs(total, k, 0, must, False) % MOD)


if __name__ == '__main__':
    main()