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 collections import defaultdict
from collections import Counter
from collections import deque
import heapq
import math
 
 
def integer():
    return int(input())
 
 
def num_lst():
    return list(map(int, input().split()))
 
 
def string():
    return input()
 
 
def str_lst():
    return list(input().split(" "))
    
def Sort(tuple):
    return(sorted(tuple, key = lambda a: a[1]))  
 

n,k,d = num_lst()
memo = defaultdict(int)

def solve(num,flag,memo):
    
    if num == 0:
        if flag:
            return 1
        return -1
    
    if num < 0:
        return -1
    
    if (num,flag) in memo:
        return memo[(num,flag)]
    
    
    s = 0
    
    for i in range(1,k+1):
        temp = solve(num -i,flag or i >= d,memo)
        if temp >= 0:
            s += temp
    
    memo[(num,flag)] = s
    
    return memo[(num,flag)]

print(solve(n,False,memo))
        