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*).
# DO NOT EDIT THIS
import functools
import math
import sys

input = sys.stdin.readline
from collections import deque, defaultdict
import heapq
import bisect
def counter(a):
    c = defaultdict(lambda : 0) # way faster than Counter
    for el in a:
        c[el] += 1
    return c

def inp(): return [int(k) for k in input().split()]
def si(): return int(input())
def st(): return input()

mxi = 10**10
mod = 10**9 + 7

# DO NOT EDIT ABOVE THIS

n, k, d = inp()

# base case: remaining_weight == 0, 1 valid construction if d
# recurrence:
# for w in 1..min(k, remaining_weight):
#   total += dp(remaining_weight - w, (w >= d) | prev_d)
#


dp = [[0] * (k + 1) for _ in range(n + 1)]

@functools.lru_cache(None)
def dp(remaining_weight=n, had_big_enough=False):
    if remaining_weight == 0:
        return 1 if had_big_enough else 0

    tot = 0
    for w in range(1, min(k + 1, remaining_weight + 1)):
        tot += dp(remaining_weight - w, had_big_enough or (w >= d))
        tot %= mod

    return tot

print(dp())







