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 math

inp = input().split(" ")
n = int(inp[0])
k = int(inp[1])
d = int(inp[2])

mod_num = 1000000007

results = dict()

def compute(target: int, k:int, d:int):

    if target in results:
        return
    
    local_results = []
    
    for x in range(d, k+1):
        new_target = target - x
        
        if (new_target < 0):
            break
        elif (new_target == 0):
            local_results.append({x: 1})
        else:
            compute(new_target, k, 1)
            res = results[new_target]
            for r in res:
                a = r.copy()
                if x in a:
                    a[x] += 1
                else:
                    a[x] = 1
                local_results.append(a)
    results[target] = local_results
    
compute(n, k, d)

final_result = results[n]
res = 0
for m in final_result:
    dividend_fact = 0
    divisor = 1
    for key in m:
        dividend_fact += m[key]
        divisor *= math.factorial(m[key])
    res += math.factorial(dividend_fact)/divisor
    res %= mod_num

print(int(res))