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 sys, io, os
import math
import bisect
import heapq
import string
from collections import defaultdict,Counter,deque
from functools import lru_cache
input = sys.stdin.readline
sys.setrecursionlimit(200)
 
def I():
    return input()
 
def II():
    return int(input())
 
def MII():
    return map(int, input().split())
 
def LI():
    return list(input().split())
 
def LII():
    return list(map(int, input().split()))
 
def GMI():
    return map(lambda x: int(x) - 1, input().split())
 
def LGMI():
    return list(map(lambda x: int(x) - 1, input().split()))
 
def WRITE(out):
  return print('\n'.join(map(str, out)))
 
def WS(out):
  return print(' '.join(map(str, out)))
 
def WNS(out):
  return print(''.join(map(str, out)))

def WSNOPRINT(out):
  return ''.join(map(str, out))

'''

'''
def solve():
    n, k, d = MII()
    mod = 10**9 + 7
    cache = defaultdict(int)

    def dfs(n, y):
        if (n,y) in cache:
            return cache[(n,y)]
        if n < 0:
            return 0
        if n == 0:
            return 1 if y else 0

        ans = 0
        for i in range(1, k+1):
            ans += dfs(n-i, i >= d or y) % mod
        cache[(n,y)] = ans
        return ans % mod

    print(dfs(n,False))

solve()