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
import threading


def main():
    n , k  , d = map(int , input().split())
    memo = {}

    def dp(_sum , flag):
        if (flag , _sum) in memo:
            return memo[(flag , _sum)]
        if _sum == n and flag:
            return 1
        if _sum > n:
            return 0
        
        ans = 0
        for i in range(1 , k+1):
            ans += dp(_sum + i , i >= d or flag)
        
        memo[(flag , _sum)] = ans
        return ans
    print(dp(0 , False))

if __name__ == '__main__':
    sys.setrecursionlimit(1 << 30)
    threading.stack_size(1 << 27)

    main_thread = threading.Thread(target=main)
    main_thread.start()
    main_thread.join()
