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 cache
from sys import stdout, stdin
from time import perf_counter as now

@cache
def choose(n, k):
    if(k > (n >> 1)):
        return choose(n, n - k)
    elif(k == 0):
        return 1
    elif(k < 0):
        return 0
    else:
        return choose(n - 1, k) + choose(n - 1, k - 1)

def inp():
    return stdin.readline().strip()

def inpl(sep=" "):
    return inp().split(sep)

def intl(in_list):
    return [int(item) for item in in_list]

def strl(in_list):
    return [str(item) for item in in_list]

def conl(in_list):
    result = ""

    for item in in_list:
        result += item
    
    return result

def solve(n, k, d):
    return sum(choose(n - 1, i) for i in range(n - k, n - d + 1))

[n, k, d] = intl(inpl())

stdout.write(f"""{solve(n, k, d)}\n""")