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
from math import *
# from pprint import pprint as pprint
# from sympy.ntheory.modular import crt
import heapq
from copy import *
from collections import *
import functools
from itertools import *
from math import sqrt, ceil
import threading
import time

# fin = sys.stdin.read().strip().split("\n")

MOD = 1000000007


def solve():

    @functools.lru_cache(maxsize=None)
    def dp(i, remains, isIncluded):
        if remains == 0:
            if isIncluded:
                return 1
            return 0

        ret = 0
        for next in range(1, k + 1):
            if next > remains:
                break
            ret += dp(i + 1, remains - next, isIncluded or (next >= d))
            ret %= MOD
        return ret % MOD

    n, k, d = [int(c) for c in input().strip().split()]
    print(dp(0, n, False))


solve()

# t = int(input())
# for _ in range(t):
#     solve()
