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*).
#I = lambda: [int(i) for i in input().split()]
#import io, os, sys
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
 
 
# n = int(input())
# l1 = list(map(int,input().split()))
# n,x = map(int,input().split())
# s = input()
mod = 1000000007
# print("Case #"+str(_+1)+":",)
 
from collections import Counter,defaultdict,deque
#from heapq import heappush,heappop,heapify
import sys
import math
import bisect
def lcm(a,b):
    return ((a*b)//math.gcd(a,b))
 
 
for _ in range(1):
    n ,k ,d = map(int,input().split())
    dp1 = [0 for i in range(n+1)]
    dp1[0] = 1
    for i in range(1,n+1):
        for j in range(1,k+1):
            if j<=i:
                dp1[i] = (dp1[i-j] + dp1[i])%mod
    dp2 = [0 for i in range(n+1)]
    dp2[0] = 1
    for i in range(1,n+1):
        for j in range(1,d):
            if j<=i:
                dp2[i] = (dp2[i-j] + dp2[i])%mod

    print((dp1[n]-dp2[n])%mod)
        
    
        
    
    
        
    
    
