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 math import factorial,ceil

mod = 1e9 + 7


def count_binom(n, k):
    if (n==0 or k==0 or k>n):
        return 0
    res = factorial(n) // (factorial(k) * factorial(n - k))
    return int(res) % mod


n, k, d = map(int, input().split())
high_p = n - d - 1
low_p = int(ceil(n / k))
res = 0 if (low_p > 1) else 1
for p in range(low_p, high_p + 1):
    if (n > d * p):
        res += count_binom(n - d * p - 1, p - 1)
    else:
        res += count_binom(n - 1, p - 1)
    res -= count_binom(n - (k + 1) * p - 1, p - 1)
    res %= mod
print(int(res))