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*).

import time
import sys
from collections import *
from bisect import *
from heapq import *

input = sys.stdin.readline

is_debug = '_local_debug_' in globals()

#int input
def inp():
    return(int(input()))

#list input
def inlt():
    return(list(map(int,input().split())))

#string input
def insr():
    return input()[:-1]

#strings input
def inss(n):
    return [insr() for _ in range(n)]

#variables input
def invr():
    return(map(int,input().split()))

#matrix input
def inmt(n):
    return [inlt() for _ in range(n)]
mod = 10**9+7
class Solution(object):
    def input(self):
        # 读入数据，返回一个参数列表
        # 例如：读取一个整数n和一个字符串s
        # return (n, s)
        return invr()

    def solve(self, n, k, d):
        ans = 0
        #flag:0表示还没有出现过权值大于等于d的边，1表示出现过
        #n:剩余的权值
        #返回值：构造出n的路径数
        memo = {}
        def dp(flag, n):
            if n == 0:
                return flag
            if n < 0:
                return 0
            if (flag, n) in memo:
                return memo[(flag, n)]
            ans = 0
            for i in range(1, k + 1):
                if i >= d:
                    ans += dp(1, n - i)
                else:
                    ans += dp(flag, n - i)
            ans %= mod
            memo[(flag, n)] = ans
            return ans
        ans = dp(0, n)
        return ans

    def output(self,ans):
        print(ans)

def solve(i):
    sol = Solution()
    param = sol.input()
    if is_debug: start = time.time()
    ans = sol.solve(*param)
    sol.output(ans)
    if is_debug:
        end = time.time()
        print(f'Case{i} Time: {(end - start) * 1000} ms')

#对于只有一个样例的题目，将inp()改为1
#t = inp() 
t = 1
if is_debug: start = time.time()
for i in range(1, t + 1):
    solve(i)
if is_debug:
    end = time.time()
    print(f'Total Time: {(end - start) * 1000} ms')