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 sortedcontainers import SortedSet
from ast import Mod
from collections import OrderedDict
from io import BytesIO, IOBase
import os
import threading
import collections
from tkinter import E
from types import GeneratorType
from itertools import combinations, permutations
from functools import lru_cache
import bisect
import sys
from array import array
import heapq
from collections import Counter
import math
from collections import defaultdict
from collections import deque
from copy import deepcopy

input = sys.stdin.readline
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc


n, k, d = map(int, input().split())



dp = [[None for i in range(2)] for z in range(n + 1)]

@bootstrap
def dfs(totalSum, state):
    if not totalSum:
        if state:
            yield 1
        yield 0;    
    
    if totalSum < 0:
        yield 0
        
    
    if dp[totalSum][state] is not None:
        yield dp[totalSum][state]    
    
    
    
    currentSum = 0
    for i in range(1, k + 1):
        newState = 1 if i >= d else 0
        newState |= state
        currentSum += (yield dfs(totalSum - i, newState))
        currentSum %= (10**9 + 7)
        
    dp[totalSum][state] = currentSum
    yield currentSum
    


print(dfs(n, 0))

        
                    
    

