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
from collections import defaultdict, deque, Counter
class Solution:
    
    def kTree(self, n, k, d):
        # TODO write an algorithm here
        self.answer = 0
        def dfs(path_sum, path):
            if path_sum == n:
                for num in path:
                    if num >= d:
                        self.answer += 1
                        break
                return self.answer
            if path_sum >= n:
                return self.answer
            
            for i in range(1, k+1):
                if i in path:
                    path[i] += 1
                else:
                    path[i] = 1
                dfs(path_sum+i, path)
                if path[i] > 1:
                    path[i] -= 1
                else:
                    del path[i]

            return self.answer
        return dfs(0, defaultdict(int))
            
 
if __name__ == "__main__":
    solution = Solution()
    n, k, d = list(map(int, input().split()))
    print(solution.kTree(n, k, d))