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*).
# https://codeforces.com/problemset/problem/431/C

import math
import random
import heapq

from collections import *
from bisect import *
from time import *
from functools import *

mod = 1000000007
memo = [[-1 for _ in range(2)] for _ in range(101)]

def solve(n, k, d, flag=0):
	if memo[n][flag] != -1:
		return memo[n][flag]

	if n == 0 and flag == 1:
		return 1

	if n <= 0:
		return 0

	count = 0
	for i in range(1, k+1):
		# flag = 1 if flag == 1 or i >= d else 0
		count += solve(n-i, k, d, 1 if flag == 1 or i >= d else 0) % mod

	memo[n][flag] = count % mod

	return memo[n][flag]

if __name__ == "__main__":
	n, k, d = list(map(int, input().split(" ")))

	print(solve(n, k, d))