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 collections import defaultdict, deque
import heapq
import sys
import threading
from functools import lru_cache

def inlt():
    return(list(map(int,input().split())))
def solve(n,k,d):
    @lru_cache(None)
    def dp(used,total):
        if total == n:
            return 1 if used else 0
        if total > n:
            return 0
        res = 0
        for i in range(1,k + 1):
            if i >= d:
                res += dp(True,total + i)
            else:
                res += dp(used,total + i)
        return res
    return dp(False,0)
def main():
    print(solve(*inlt()))
threading.stack_size(1 << 27)
sys.setrecursionlimit(1 << 30)
main_thread = threading.Thread(target=main)
main_thread.start()
main_thread.join()